Answer:
Explanation:
For the Magnesium Ion, ([tex]Mg^{+2}[/tex])
Electrons: 10
Protons: 12
Neutrons: 12
Use your periodic table to answer the following question.
A scientist was looking at two elements, Titanium (Ti) and Uranium (U). Which observation would be a true statement?
Question 7 options:
Uranium is larger than Titanium and it has higher electronegativity
Titanium is larger than Uranium and it has a higher electronegativity
Uranium is larger than Titanium and it has a higher ionization energy
Titanium is smaller than Uranium and has a higher ionization energy
Answer:
Option B: Uranium is larger than Titanium and it has a higher ionization energy
Explanation:
From left to right across the periodic table, electronegativity increases.
So, going from left to right across the periodic table, Titanium comes before uranium. Thus;
Uranium will have a higher electronegativity.
So option A is false
From left to right across the periodic table, ionization energy increases.
Thus;
Uranium will have a higher ionization energy than Titanium.
Thus; option B is correct.
what is a a metamorphic rock
Answer:
A metamorphic rock is rock that comes from mountains and its little rocks squished together
Explanation:
What is the difference between mass number and average atomic mass?
(choose all that apply)
A: Mass number is a whole number, average atomic mass is a decimal number
B: Mass number is protons + neutrons, average atomic mass is all the different protons + neutrons
C: Mass number is the mass of one element, average atomic mass is the number of
protons
D: There is no difference
Answer:
the answer is c. I think
Explanation:
_________energy from the sun
Answer:
radiant energy, solar energy
During nuclear fusion, radiant energy and solar energy from the core of the Sun is released. Hope it helps!
Suppose a 20.0 g gold bar at 35.0°C absorbs 70.0 calories of heat energy. Given that the specific heat of gold is 0.0310 cal/g °C, what is the final
temperature of the gold bar?
We know, change in temperature is given by :
[tex]T_2-T_1=\dfrac{q}{mC_p(Gold)}[/tex]
Putting all given values, we get :
[tex]T_2-T_1=\dfrac{70\ cal}{20\ g\times 0.0310\ cal/g^o\ C}\\\\T_2-T_1=112.90^oC\\\\T_2-35^oC=112.90^oC\\\\T_2=(112.90+35)^oC\\\\T_2=147.9^oC[/tex]
Hence, this is the required solution.
Where are the instructions that tell the cell everything it needs to survive?
mitochondrion
nucleolus
endomembrane system
ribosome
Answer:
Nucleolus
Explanation:
It holds the DNA
Answer:
Nucleolus
Explanation:
I did the test. You’re welcome. ✌️
You have a mixture of salt, sand, iron fillings, and large rocks in a big bucket. Your goal is to separate the four different items into their own bucket. Write a detailed procedure showing the step by step process on to separate each item.
Answer:Please see answers in the explanation column
Explanation:
In a mixture of salt, sand, iron fillings, and large rocks in a big bucket.
The process of separation is as follows:
Ist step ---
Use of magnet : Because Irons are attracted to magnet, Introducing a magnet to the mixture will cause the iron fillings to cling to the magnet leaving the other constituents behind
2nd Step
we eliminate the large rocks by carefully Sorting or picking them out since they are large and visible or we use a MESH or Sieve and carry out the separation by pouring the rock, salt and sand mixture on top of the mesh with continuous shaking so as to filter the smaller particles,The smaller particles will pass through the mesh while the bigger rocks remains on top.
3rd Step
We are left with the salt and sand. we will then introduce water to mix the salt and sand. salt dissolves completely in water while the sand settles as the bottom of the container. We can then separate the sand by decantation and then filtration to remove any left over sand.
4th step
Evaporation is then carried out on the salt solution to separate the salt from water.
Which of the following observations is usually not evidence of a chemical change ?
A. Change of shape
B . Formation of a precipitate
C.Giving off of a gas
D.giving off heat and/or light
Answer:
D
Explanation:
Have a nice day
Among the following observations which is usually not evidence of a chemical change is change of a shape.
What is chemical change?Chemical changes will occur when the composition of the substance is changes internally or converted into any new product.
Change of shape is an example of physical change as it is not essential if shape changes then chemical properties also changes.Formation of a precipitate is a chemical process because in this process any new substance is formed.Giving off a gas is also a chemical process as in this change an extra product is released.Giving off heat or light is also a chemical change because this heat will be calculated as a energy.So, change of shape is not a chemical change.
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2. A gas has volume of 4.2L at 110 kPa. If temperature is constant, find the pressure of gas when the volume is change to 11.3L.
Answer:
40.88 kPa.
Explanation:
Given that,
Volume, V₁= 4.2 L
Pressure, P₁ = 110 kPa
We need to find the pressure of the gas when the volume is changed to 11.3L at constant temperature.
According to Boyle's law, at constant temperature, volume is inversely proportional to the pressure. Mathematically,
[tex]P_1V_1=P_2V_2\\\\P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{4.2\ L\times 110\ kPa}{11.3\ L}\\\\P_2=40.88\ kPa[/tex]
So, the new pressure is 40.88 kPa.
How a cation is formed by an atom?
Answer:
Cations are positively charged atoms and hence we need to make the atom positively charged in order to get a cation
We know that an atom is neutral as a whole, so we have equal number of electrons and protons
since we cannot mess with the number of protons in an atom, we have to do it by altering the number of electrons
If we reduce the amount of electrons in an atom, the net charge will be positive and hence a cation will be formed
The most common form of elemental sulfur is S8, in which eight sulfur atoms linked in a ring of single bonds. At high temperature, in the gas phase, S8 can break apart to give S2, the sulfur analog of molecular oxygen:
S8(g) 4S2(g) ΔH = +239 kJ at T = 800 K
Using the appropriate data in Table 3-2 S-S 240 calculate the S=S double-bond energy (in kJ/mol) in S2(g).
Answer:
Explanation:
In S₈ , there are 8 single bonds which breaks up first . Energy absorbed
= 8 x 240 = 1920 kJ
In S₈ four double bonds of S₂ are formed . Let bond energy be x . In this process energy will be released . energy released in four S₂ molecules formed = 4 x
Given
1920 + 4x = 239
4x = 239 - 1920
x = - 420.25 kJ .
So bond energy of S₂ = 420.25 kJ .
Cl2 reacts with the element Sr to form an ionic compound. Based on periodic properties, identify a molecule, X2, that is likely to to react with Sr in a way similar to how Cl2 reacts with Sr. Justify your choice.
Answer:
The bromine molecule, Br2 will likely react with Sr in a similar way to chlorine molecule, Cl2.
Explanation:
Chlorine belongs to group 7 of the periodic table. Elements in this group are known as halogens. Its molecule has the formula Cl2. The next element in the group after chlorine is bromine. Its molecule has a formula Br2. It has similar properties as chlorine, therefore, it would react with Strontium in a similar way to chlorine.
Elements belonging to the same group of the periodic the table have similar chemical properties as they contain the same number of valence electrons. Chlorine and Bromine both belong to group 7 of the periodic table and each have seven valence electrons. They both react with metals to form salts even though reaction with chlorine is more vigorous as it is more reactive than bromine. They both form negatively-charged ions with a charge of -1.
Reaction of Strontium with the halogens:
With chlorine: Sr + Cl2 ---> SrCl2
With bromine: Sr + Br2 ---> SrBr2
The salts formed are both crystalline salts with high melting and boiling points.
A molecule X₂ that will likely react with Sr in a way similar to how Cl₂ reacts with Sr is I₂
Strontium, Sr is a group 2 element.
Chlorine, Cl is a group 17 element. All members of this group have 7 valence electrons and as such, they have similar chemical properties. Other members of this group include fluorine (F), bromine (Br), iodine (I).
Chlorine molecule (Cl₂) reacts with strontium (Sr) as follow:
Cl₂ + Sr —> SrCl₂Iodine molecule (I₂) will also reacts in the same way since they are in the same group.
I₂ + Sr —> SrI₂Learn more: https://brainly.com/question/4725757
What is the difference between a physical change and a chemical change?
Explanation:
Hey there!
The differences between physical chamge and chemical change are;
The change is the temporary change in which no new substances are formed. But chemical change are the change in which new substances are formed. Physical change are reversible. But chemical change are irreversible. No new substances are formed with new properties in physical change. But in chemical change new substances with new properties are formed.Hope it helps...
Help me find the answer please
Answer:
1. A
2. C
3. B
4. G
5. H
6. D
7. F
8. E
Explanation:
Can someone please help?
Answer: The correct option is, (a) 9.0 g
Explanation : Given,
Mass of [tex]N_2H_4[/tex] = 8.0 g
Mass of [tex]N_2O_4[/tex] = 92 g
Molar mass of [tex]N_2H_4[/tex] = 32 g/mol
Molar mass of [tex]N_2O_4[/tex] = 92 g/mol
First we have to calculate the moles of [tex]N_2H_4[/tex] and [tex]N_2O_4[/tex].
[tex]\text{Moles of }N_2H_4=\frac{\text{Given mass }N_2H_4}{\text{Molar mass }N_2H_4}[/tex]
[tex]\text{Moles of }N_2H_4=\frac{8.0g}{32g/mol}=0.25mol[/tex]
and,
[tex]\text{Moles of }N_2O_4=\frac{\text{Given mass }N_2O_4}{\text{Molar mass }N_2O_4}[/tex]
[tex]\text{Moles of }N_2O_4=\frac{92g}{92g/mol}=1mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]2N_2H_4(g)+N_2O_4(g)\rightarrow 3N_2(g)+4H_2O(g)[/tex]
From the balanced reaction we conclude that
As, 2 mole of [tex]N_2H_4[/tex] react with 1 mole of [tex]N_2O_4[/tex]
So, 0.25 moles of [tex]N_2H_4[/tex] react with [tex]\frac{0.25}{2}=0.125[/tex] moles of [tex]N_2O_4[/tex]
From this we conclude that, [tex]N_2O_4[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]N_2H_4[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]H_2O[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]N_2H_4[/tex] react to give 4 mole of [tex]H_2O[/tex]
So, 0.25 mole of [tex]N_2H_4[/tex] react to give [tex]\frac{4}{2}\times 0.25=0.5[/tex] mole of [tex]H_2O[/tex]
Now we have to calculate the mass of [tex]H_2O[/tex]
[tex]\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O[/tex]
Molar mass of [tex]H_2O[/tex] = 18 g/mole
[tex]\text{ Mass of }H_2O=(0.5moles)\times (18g/mole)=9.0g[/tex]
Therefore, the mass of [tex]H_2O[/tex] produced is, 9.0 grams.
If methane undergo combustion, what mass of methane must be burnt to produce 27.5 g of CO2?
Answer:
Mass of methane burnet 10.025 g.
Explanation:
Given data:
Mass of CO₂ produced = 27.5 g
Mass of methane burnet = ?
Solution:
Balanced chemical equation.
CH₄ + 2O₂ → CO₂ + 2H₂O
First of all we will calculate the number of moles of CO₂ produced.
Number of moles = mass/ molar mass
Number of moles = 27.5 g/ 44 g/mol
Number of moles = 0.625 mol
Now we will compare the moles of CO₂ with CH₄.
CO₂ : CH₄
1 : 1
0.625 : 0.625
0.625 moles of CH₄ are burnet.
Mass of methane burned:
Mass = number of moles × molar mass
Mass = 0.625 mol × 16.04 g/mol
Mass = 10.025 g
ou are performing a ton test for household ammonia. you place 150 ml of distilled water in a 500 ml beaker first. you added 1 ml of the household ammonia and add distilled water to top up a total volume of 200 ml, the odor can be detected in this dilution. take 50 ml of the above diluted sample as an intermediate sample and continually dilute it to 200 ml and the odor is still detectable. you decided to reduce to 10 ml of the first time diluted household ammonia and the odor was just detectable. how much is the ton of the household ammonia
Answer: the ton of the household ammonia is 4000
Explanation:
Given that;
the initial ammonia in 200mL sample = 1ml
also decided to reduce to 10 ml of the first time diluted household ammonia and the odor was just detectable
10ml ammonia content = 1/200 * 10 = 1/20ml
we know that total volume is 200 ml
therefore the TON of household ammonia will be = total volume / volume of ammonia
TON = total volume / volume of ammonia
we substitute
TON = 200 / (1/20)
TON = 200 / 0.05
TON = 4000
therefore the ton of the household ammonia is 4000
What is the volume of the water in this graduated cylinder?*
A car travels at a speed of 54 km/hr. How many meters will it travel in 1 second?
Answer: 15 METERS IN A SECOND
Explanation: 54km = 54000m 54000m / 60 = 900 900/60 = 15
Calculate the volume of an object (in mL) if the density of the object is 4.08 g/mL and the mass of the object is 10.02 g.
Round your answer to one decimal.
Answer:
The answer is 2.5 mLExplanation:
The volume of a substance when given the density and mass can be found by using the formula
[tex]volume = \frac{mass}{density} \\ [/tex]
From the question
mass of object = 10.02 g
density = 4.08 g/mL
The volume of the object is
[tex]volume = \frac{10.02}{4.08} \\ = 2.4558823...[/tex]
We have the final answer as
2.5 mL to one decimal placeHope this helps you
Determine the number of neutrons, protons, and electrons for a bromide ion that has a mass number of 76, and a charge of
-1.
Answer:
4 5 6
Explanation:
Atomic models have changed over the decades. Two early atomic models can be seen here. There is a dramatic change in the models, as Rutherford experimented with the cathode ray tube and charged particles. Differentiate between the two models.
Question 14 options:
A) Rutherford's model shows negative charges dispersed throughout the atom.
B) Rutherford's model shows negative particles orbiting the central nucleus.
C) Rutherford's model shows the positive charge of an atom as a very small area.
D) Thomson's model shows at sea of negative charged particles surrounding a small, positive area.
Answer:
C. Rutherford's model shows the positive charge of an atom as a very small area.
Explanation:
Rutherford's model shows the positive charge of an atom as a very small area.
The Thompson model of the atom depicted the atom as a sphere of positively charges into which negative charges were embedded. This is called the plum - pudding model of the atom.
The Rutherford's model of the atom depicted the atom as having a core containing positive charges. This core occupies a very small area and is surrounded by electrons. This is called the planetary model.
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What mass (in g) of nitrogen are needed to react completely with 5.8g of hydrogen?
[tex]N_{2}[/tex] + [tex]3H_{2}[/tex] —> [tex]2NH_{3}[/tex]
Answer:
Explanation:
mass of N2 Mass of3 H2
2x (14) g/mole 3x(2x1) g/mole
X g 5.8 g
mass of nitrogen = 27 g
What is needed to remove a phosphate from ATP?
Answer:
A water molecule
Explanation:
The removal of a phosphate molecule from ATP to form ADP is a hydrolysis reaction. Hydrolysis reactions require water and release energy.
The converse reaction - when a phosphate is added to a molecule of ADP to form ATP - releases water and requires energy.
Use the periodic table in the tools bar to complete the electron configurations for the following elements:
Lithium (Li): 1sA2sB
A =
B =
Answer:
A= 2, B= 1
Explanation:
In the given electronic configuration of Lithium, the values of A and B are 2 and 1, respectively.
The electronic configuration of an atom describes how its electrons are distributed among the various energy levels and orbitals. It is typically represented using a series of numbers and symbols that indicate the number of electrons in each orbital.
To complete the electron configuration for lithium (Li), we'll refer to the periodic table:
The atomic number of lithium is 3, which means it has 3 electrons. The electron configuration for lithium can be determined as follows:
[tex]1s^2 2s^1.[/tex]
In this case,
the first energy level (n = 1) contains 2 electrons ( represented as [tex]1s^2[/tex]), and the second energy level (n = 2) contains 1 electron ( represented as [tex]2s^1)[/tex].So, for lithium (Li), the completed electron configuration is:
[tex]1s^2 2s^1.[/tex]
Therefore, the values of A and B in the given electronic configuration of Lithium are A = 2, B = 1, respectively.
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If you lived in Flagstaff, Arizona, how much salt (NaCl) would you have to add to your spaghetti water to get it to boil at 100 oC. Assume you are using 2 quarts of water in your spaghetti pot. Report the amount of salt in units you might actually be able to measure in your kitchen. If you have a scale that shows grams that will be okay. (Hint: To get started find out the actual boiling point of water in Flagstaff.) It is at an elevation above 7000 ft.)
Answer:
Explanation:
This question is both theoretical and practical. While the theoretical aspect will be detailed fully here, the practical aspect will be provided as a form of guidance.
Water generally boils at 100°C when altitude (in feet) is 0. One of the colligative properties that occurs when salt is added to water is that there is a boiling point elevation(meaning an increase in boiling point). For instance, if 20g of salt is added to about 5.3 quarts of water, the boiling point of water will increase from 100°C to 100.04°C.
However, when the altitude/elevation of a place is about 7000 ft (like in Flagstaff, Arizona), water will boil at 95.3°C. In order to get 2 quarts of water to boil at 100°C in Flagstaff;
20g causes an increase in boiling point by 0.04°C (100°C to 100.04°C) in 5.3 quarts of water
What gram will increase the boiling point by same 0.04°C in 2 quarts
20g ⇒ 5.3
X ⇒ 2
5.3 X ⇒ 40g
X = 40 ÷ 5.3
X = 7.55g
Hence, 7.55g will cause an increase in boiling point by 0.04°C (from 100°C to 100.04°C) in 2 quarts of water
What mass of salt will increase the boiling point by 4.7°C (95.3°C to 100°C)
7.55g ⇒ 0.04
X ⇒ 4.7
X × 0.04 ⇒ 7.55 × 4.7
0.04X ⇒ 35.5
X = 887.5g
Hence, in order for the spaghetti water to boil at 100°C, 887.5g of salt needs to be added.
For the practical part of the question, some Kitchen scales have an accuracy of .25kg (250g) and some have an accuracy of .2 kg (200g) and some have an accuracy of .5kg (500g). The one your kitchen has will determine the amount of salt that you can measure. For example, if your kitchen scale/balance has an accuracy of 250g/0.25kg, then you can only measure 750g of the 887.5g (as the rest is 137.5g, which is not up to 250g of the scale's accuracy) of the required salt measurement. However, if you have a digital balance that can measure up to 2kg/2000g in one decimal place, that's the perfect balance to measure this salt.
How many formula units make up 36.2 g of magnesium chloride (MgCl2)?
Answer:
2.29 × 10²³ formula units of MgCl₂
Explanation:
Given data:
Mass of MgCl₂ = 36.2 g
Number of formula units = ?
Solution:
The given problem will be solved by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water = 6.022 × 10²³ formula units of water.
In 36.2 g of MgCl₂:
Number of moles = mass / molar mass
Number of moles = 36.2 g / 95.211 g/mol
Number of moles = 0.38 mol
1 mole = 6.022 × 10²³ formula units
0.38 mol / 1mol × 6.022 × 10²³ formula units
2.29 × 10²³ formula units of MgCl₂
Gallium (Ga) consists of two naturally occurring isotopes with masses of 68.926 and 70.925 amu.
Part A. How many protons and neutrons are in the nucleus of isotope with mass of 68.926 amu? Express your answers as an integers. Enter your answers numerically separated by a comma. p, n = ______.
Part B. How many protons and neutrons are in the nucleus of isotope with mass of 70.925 amu? Express your answers as an integers. Enter your answers numerically separated by a comma. p, n = _______.
Part C. Write the complete atomic symbol for each, showing the atomic number and mass number. Express your answers as isotopes separated by a comma.
Answer:
Part A. p, n = 31, 38.
Part B. p, n = 31, 40.
Part C. [tex]^{69}_{31}Ga,\ ^{71}_{31}Ga[/tex]
Explanation:
Hello,
Part A.
In this case, since the atomic number for gallium is 31, we can say that the first isotope has also 31 protons and the neutrons are computed by using its molar mass as a whole number (69):
[tex]neutrons=mass-protons=69-31=38[/tex]
Thus, result is p, n = 31, 38.
Part B.
In this case, since the atomic number for gallium is 31, we can say that the second isotopes has also 31 protons and the neutrons are computed by using its molar mass as a whole number (71):
[tex]neutrons=mass-protons=71-31=40[/tex]
Thus, result is p, n = 31, 40.
Part C.
Finally, the isotopes as shown as:
[tex]^{69}_{31}Ga,\ ^{71}_{31}Ga[/tex]
Best regards.
VSEPR theory predicts that an atom with one lone pair and three bonding pairs (such as the N-atom in aniline) will have a tetrahedral electron geometry and a trigonal pyramidal molecular geometry due to steric repulsions between H-atoms and the N-atom lone pair. However, in question 5 you observed that the N-atom in aniline is not perfectly sp3 hybridized (i.e. the hybridization is different from that predicted for a tetrahedral electron geometry). Briefly describe all of the factors that result in the calculated hybridization of the N-atom lone pair
Answer: The lone pair of electron on nitrogen is accommodated in a 2p orbital hence it interacts with the pi system in aniline.
Explanation:
Aniline is less basic than amines. This is because, the nitrogen atom in aniline is not purely sp3 hybridized. Its actual hybrization state is closer to sp2 because the lone pair on nitrogen is accommodated in a 2p orbital. The nitrogen atim in aniline is planar and its
lonely pair interacts with the pi electron system of aniline. This makes the lone pair unavailable for protonation hence aniline is less basic than amines.
Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.80 g of sodium carbonate is mixed with one containing 5.43 g of silver nitrate.
Part A)
How many grams of sodium carbonate are present after the reaction is complete?
Part B)
How many grams of silver nitrate are present after the reaction is complete?
Part C)
How many grams of silver carbonate are present after the reaction is complete?
Part D)
How many grams of sodium nitrate are present after the reaction is complete?
Answer:
Part A) [tex]m_{Na_2CO_3}^{leftover}=3.74g[/tex]
Part B) Nothing as it is the liming reactant.
Part C) [tex]m_{Ag_2CO_3}=4.42gAg_2CO_3[/tex]
Part D) [tex]m_{NaNO_3} =2.72gNaNO_3[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]Na_2CO_3(aq)+2AgNO_3(aq)\rightarrow Ag_2CO_3(s)+2NaNO_3(aq)[/tex]
Thus, in order to proceed, we first need to identify the limiting reactant is it yielding the least moles of silver carbonate for instance, thus, for the given mass of sodium carbonate (molar mass 106 g/mol) and silver nitrate (molar mass 170 g/mol) we obtain for silver carbonate:
[tex]n_{Ag_2CO_3}^{by\ Na_2CO_3}=3.80gNa_2CO_3*\frac{1molNa_2CO_3}{106gNa_2CO_3}*\frac{1molAg_2CO_3}{1molNa_2CO_3} =0.0359molAg_2CO_3\\\\n_{Ag_2CO_3}^{by\ AgNO_3}=5.43gAgNO_3*\frac{1molAgNO_3}{170gAgNO_3}*\frac{1molAg_2CO_3}{2molAgNO_3} =0.0160molAg_2CO_3[/tex]
In such a way, we can see that the silver nitrate yields less amount of silver carbonate, that is why it is the limiting reactant. Thus, we answer:
Part A) Since sodium carbonate is the excess reactant, we need to compute the consumed grams by the limiting silver nitrate as shown below:
[tex]m_{Na_2CO_3}^{consumed}=5.43gAgNO_3*\frac{1molAgNO_3}{170gAgNO_3}*\frac{1molNa_2CO_3}{2molAgNO_3} *\frac{106gNa_2CO_3}{1molNa_2CO_3} =1.69gNa_2CO_3[/tex]
Therefore, the leftover of sodium carbonate is:
[tex]m_{Na_2CO_3}^{leftover}=5.43g-1.69g=3.74g[/tex]
Part B) Since silver nitrate is the limiting reactant no leftover are present after the reaction as it was all consumed.
Part C) As the 5.43 g of silver nitrate yield 0.0160 moles of silver carbonate, the corresponding mass is computed as shown below:
[tex]m_{Ag_2CO_3}=0.0160molAg_2CO_3*\frac{276gAg_2CO_3}{1molAg_2CO_3}=4.42gAg_2CO_3[/tex]
Part D) Finally, via stoichiometry, we compute the grams of sodium nitrate that are yielded:
[tex]m_{NaNO_3}=5.43gAgNO_3*\frac{1molAgNO_3}{170gAgNO_3}*\frac{2molNaNO_3}{2molAgNO_3} *\frac{85gNaNO_3}{1molNaNO_3} =2.72gNaNO_3[/tex]
Best regards.