Pb(NO3)2+K2CrO4=PbCrO4+KNO3 reaction type
The reaction between Pb(NO₃)₂ (lead(II) nitrate) and K₂CrO₄ (potassium chromate) can be classified as a double displacement or precipitation reaction.
In this reaction, the lead(II) ion (Pb²⁺) from Pb(NO3)2 combines with the chromate ion (CrO₄²⁻) from K₂CrO₄ to form lead(II) chromate (PbCrO₄). Meanwhile, the potassium ion (K⁺) from K₂CrO₄ combines with the nitrate ion (NO₃⁻) from Pb(NO₃)₂ to form potassium nitrate (KNO₃).
The balanced chemical equation for the reaction is:
Pb(NO₃)₂ + K₂CrO₄ → PbCrO₄ + 2KNO₃
In this equation, the subscripts and coefficients are used to balance the number of atoms on each side of the equation to satisfy the law of conservation of mass.
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How many grams of Ca ( OH )2 , ( Mwt = 74 g / mol ) exists in 2500 mL of 0.0250 M Ca ( OH )2 ?
Answer:
4.6g
Explanation:
We are given the following parameters:
Molecular weight of the compound = 74g/mol
Volume = 2500mL = 2.5L or 2.5dm³
Concentration = 0.025M
Unknown:
Mass of compound = ?
Solution:
To solve this problem, we need to find the number of moles from the given concentration and volume values.
Number of moles = concentration x volume
So;
Number of moles = 0.025 x 2.5 = 0.0625mol
Therefore, mass of compound;
Mass = number of moles x molar mass
Mass = 0.0625 x 74 = 4.6g
Nitrogen forms more oxides than any other element. The percents by mass of N in three different nitrogen oxides are (I) 46.69%; (II) 36.85%; (III) 25.94%. For each compound, determine (a) the simplest whole-number ratio of N to O, and (b) the number of grams of oxygen per 1.00 g of nitrogen.
Answer:
[1]. NO; 1.14 g,
[2]. N₂O₃; 1.71 g,
[3]. N₂O₅; 2.86 g.
Explanation:
[1]. For the first oxide, there is 46.69% of Nitrogen and [100 - 46.69%] of Oxygen, that is 53.31% of oxygen. The simplest whole-number ratio of N to O = [46.69/14.01] ÷ [53.31/ 16] = 3.33/3.33 = 1 : 1.
Hence, Nitrogen =1 and oxygen is 1, the compound = NO.
The number of grams of oxygen per 1.00 g of nitrogen for this compound is calculated as below:
The number of grams of oxygen per 1.00 g of nitrogen for this compound = (1/ 0.4669) - 1 = 2.14 - 1 = 1.14 g.
[2]. For the second oxide, there is 36.85% of Nitrogen and [100 - 36.85%] of Oxygen, that is 63.15% of oxygen. The simplest whole-number ratio of N to O = [36.85/14.01] ÷ [63.15/ 16] = 2.63/3.94 = 1 : 1.5.
Hence, Nitrogen =1 and oxygen is 1.5. There is need to change it to whole number, therefore, when it is multiplied by 2, it gives 2:3. Therefore, the compound = N₂O₃.
The number of grams of oxygen per 1.00 g of nitrogen for this compound is calculated as below:
The number of grams of oxygen per 1.00 g of nitrogen for this compound = (1/ 0.3685) - 1 = 2.71 - 1 = 1.71 g.
(3). For the third oxide, there is 25.94% of Nitrogen and [100 - 25.94%] of Oxygen, that is 74.06% of oxygen. The simplest whole-number ratio of N to O = [25.94/14.01] ÷ [74.06/ 16] = 1.85/4.63= 1 : 2.5.
Hence, Nitrogen =1 and oxygen is 2.5. There is need to change it to whole number, therefore, when it is multiplied by 2, it gives 2:3. Therefore, the compound = N₂O₅.
The number of grams of oxygen per 1.00 g of nitrogen for this compound is calculated as below:
The number of grams of oxygen per 1.00 g of nitrogen for this compound = (1/ 0.2594) - 1 = 3.86 - 1 = 2.86 g
Please help and try to explain if you can!
Answer: 4.76 * 10^2
476.
you bring the ending decimal to the first number and it takes you two hops to get the first number so that's what you put in the box after the 10
NEED HELP!
C=46.67%, H=4.48%, N=31.10%, O=17.76%.
The molecular weight is 180.16g/mol.
Answer:
[tex]C_7H_8N_4O_2[/tex]
Explanation:
Hello!
In this case, since the determination of an empirical formula is covered by first computing the moles of each atom as shown below:
[tex]n_C=\frac{46.47g}{12g/mol}=3.9mol\\\\ n_H=\frac{4.48g}{1g/mol} =4.5mol\\\\n_N=\frac{31.10g}{14g/mol} =2.2mol\\\\n_O=\frac{17.76g}{16g/mol} =1.1mol[/tex]
Now, we divide each moles by the fewest moles (those of oxygen), to obtain the subscripts in the empirical formula:
[tex]C:\frac{3.9}{1.1}=3.5 \\\\H:\frac{4.5}{1.1}=4 \\\\N:\frac{2.2}{1.1} =2\\\\O:\frac{1.1}{1.1} =1[/tex]
Thus, the empirical formula, taken to the nearest whole subscript is:
[tex]C_7H_8N_4O_2[/tex]
Whose molar mass is 180.16, therefore the empirical formula is the same to the molecular one.
Best regards!
Use the atomic masses of the elements to determine the percent composition by mass of sodium and chlorine in sodium chloride.
Answer:
Explanation:
Percent composition of sodium chloride .
NaCl
Mol weight = 23 + 35.5 = 58.5 g
58.5 g of NaCl contains 23 g of Na
100 g of NaCl contains 23 x 100 / 58.5 g of Na
= 39.3 %
Na = 39.3 %
58.5 g of NaCl contains 35.5 g of Cl
100 g of NaCl contains 35.5 x 100 / 58.5 g of Na
= 60.7 %
Cl = 60.7 % .
19. Which of the following has 3 significant figures? 0.0730 O 300 4003
Answer:
0.0730
Explanation:
decimal 0s don't count except for ones at the end
Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa gage, and the atmospheric pressure is 100 kPa absolute. Some time later, after considerably more gas is pumped into the tank, the pressure in the tank is 212 kPa gage, and the temperature is still 13°C. What will be the ratio of the mass of natural gas in the tank when p = 212 kPa gage to that when the pressure was 117 kPa gage?
For this situation in which the tank volume is the same before and after filling, which of the following is the correct relation for the ratio of the mass after filling M2 to that before filling M1 in terms of gas temperatures T1 and T2 and pressures p1 and p2?
a. M2/M1= p2T2/p1T1
b. M2/M1= p1T2/p2T1
c. M2/M1= p2T1/p1T2
d. M2/M1= p1T1/p2T2
1. What is the absolute pressure in the tank before filling?
2. What is the absolute pressure in the tank after filling?
3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?
Answer:
1. the absolute pressure in the tank before filling = 217 kPa
2. the absolute pressure in the tank after filling = 312 kPa
3. the ratio of the mass after filling M2 to that before filling M1 = 1.44
The correct relation is option c ([tex]\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }[/tex])
Explanation:
To find -
1. What is the absolute pressure in the tank before filling?
2. What is the absolute pressure in the tank after filling?
3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?
As we know that ,
Absolute pressure = Atmospheric pressure + Gage pressure
So,
Before filling the tank :
Given - Atmospheric pressure = 100 kPa , Gage pressure = 117 kPa
⇒Absolute pressure ( p1 ) = 100 + 117 = 217 kPa
Now,
After filling the tank :
Given - Atmospheric pressure = 100 kPa , Gage pressure = 212 kPa
⇒Absolute pressure (p2) = 100 + 212= 312 kPa
Now,
As given, volume is the same before and after filling,
i.e. [tex]V_{1}[/tex] = [tex]V_{2}[/tex]
As we know that, P ∝ M
⇒ [tex]\frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }[/tex]
⇒[tex]\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }[/tex]
⇒[tex]\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378[/tex] ≈ 1.44
Now, as we know that PV = nRT
As V is constant
⇒ P ∝ MT
⇒[tex]\frac{P}{T}[/tex] ∝ M
⇒[tex]\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }[/tex]
So, The correct relation is c option.
What part of the underside of a gecko's feet
allows it to stick to surfaces?
O cilia
O heat
O pillars
O webbing
Answer:
it was pillars
Explanation:
i guessed and it was right LOL
Answer:
O pillars
Explanation:
Right on EDGE 2021
When fat comes in contact with sodium hydroxide, it produces soap and glycerin. Determine whether this is a physical change or a chemical change. Explain your
Answer:
This is a chemical change because new substances are formed with different properties and identities.
Determine the mass of oxygen that is needed to react with 33 grams of phosphorus.
4P + 5 O2 → 2 P205
A
27.3 grams O2
B
1.33 grams O2
С
213 grams 02
D
42.6 grams 02
Answer:
D = 42.6 g
Explanation:
Given data:
Mass of oxygen needed = ?
Mass of phosphorus react = 33 g
Solution:
Chemical equation:
4P + 5O₂ → 2P₂O₅
Number of moles of phosphorus:
Number of moles = mass/molar mass
Number of moles = 33 g/ 30.97 g/mol
Number of moles = 1.065 mol
now we will compare the moles of P and oxygen.
P : O₂
4 : 5
1.065 : 5/4×1.065 = 1.33mol
Mass of oxygen needed:
Mass = number of moles × molar mass
Mass = 1.33 mol × 32 g/mol
Mass = 42.6 g
Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you the substance is a pure substance or a mixture, if you can.
Sample A is of a coarse grey powder with a faint unpleasant smell. of the powder is put into a funnel lined with a sheet of thick paper. Distilled water is poured slowly over the powder. Most of the powder disappears, but of a gritty black sand-like material is left on the surface of the paper. Pouring more water over the black material doesn't change how much of it there is.
Sample B is a solid yellow cube with a total mass of 50.0 g. The cube is put into a beaker filled with 250. mL of water. The cube collapses into a small pile of orange powder at the bottom of the beaker. When this powder is filtered out, dried and weighed, it has a total mass of 29.9 g. If the experiment is repeated with 500. mL of water, the powder that's left over has a mass of 10.0 g.
Answer:
Sample A - mixture
Sample B- Mixture
Explanation:
Looking at sample A, we can see that as water was poured over sample A, the sample was separated into its components as the powder disappeared leaving behind a gritty black sand-like material on the surface of the paper. A separation of the mixture has taken place.
In sample B, we can clearly see that it is a mixture because the amount of solid recovered is much less than the total mass of the solid put into the beaker. The sample must have been separated into its components.
How (and where) does oceanic lithosphere (crust) form?
Answer:
Oceanic lithosphere forms at midocean ridges, where hot magma upwells, and then cools to form plates as the material moves away from the spreading center. As the plate cools, heat flow decreases and the seafloor deepens
Explanation:
Brainliest plzzzzzzz
A scooter maintains a pace of 1.45 minutes per mile, with fuel economy of 85 miles per gallon. How many seconds can it travel on 145 mL of fuel under these conditions
Answer:
281s
Explanation:
Given parameters:
Speed of the scooter = 1.45min per mile
Fuel economy of engine = 85miles per gallon
Quantity of fuel = 145mL
Unknown:
Time of travel with the volume of fuel given = ?
Solution:
To solve this problem, we need to find the distance the fuel will last.
Rate of fuel consumption by the engine = 85miles per gallon
Convert 145mL to gallons;
3785.41mL = 1 gallon
145mL will therefore give [tex]\frac{145}{3785.41}[/tex] = 0.038gallons
So;
Distance covered = 85miles per gallon x 0.038gallons = 3.23miles
From;
Rate of travel = [tex]\frac{time }{distance}[/tex]
Time = rate of travel x distance = 1.45 minutes per mile x 3.23miles
Time = 4.7min
1 min = 60s
4.7min = 4.7 x 60 = 281s
1. 2.0 mL/s x 4.0 hrs = ? L
How do you do this conversion problem
Answer:
29 L.
Explanation:
Hello!
In this case, considering that we are performing a conversion by which the time should be cancelled out to obtain liters, we first need to convert the seconds on bottom to hours and then the volume on top to liters, just a shown down below:
[tex]2.0\frac{mL}{s} *\frac{1L}{1000mL} *\frac{3600s}{1hr}*4.0hr\\\\=28.8L[/tex]
Which turns out 29 L with 2 significant figures.
Best regards!
How many moles are in 25.58 grams of H20? SHOW WORK
Answer:
1.42 moles
Explanation:
25.58 g of H₂O .
Molecular weight of water = 2 x 1 + 16 = 18
1 mole of water = 18 g of water
18 g of water = 1 mole
25.58 g of water = 25.58 / 18 mole
= 1.42 moles .
What does an electric field
surround?
Answer:
An electric field is the physical field that surrounds each electric charge and exerts force on all other charges in the field, either attracting or repelling them.
Why is it necessary for the radio of atoms to be expressed as the distance between the nuclei of the two identical atoms bonded together?
Describe what an alchemist does
Why the copper is able to replace the silver, but the reverse reaction will not result in a replacement reaction?
Answer:
Copper is more reactive and higher in the activity series
Explanation:
Copper is able to replace the silver in a replacement reaction because it is more reactive and higher in the activity series.
Substances or ions that are higher in the activity series are typically more reactive than those ones below them.
In like manner, they are able to replace the lowers ones in a chemical reaction due to their higher reactivity potential.
A less reactive element and a lower one in the activity series cannot displace a higher one in the series.
How much of a 500 gram sample of potassium-42 is left after
62 hours? The half-life of K-42 is 12.4 hours?
Answer:
The correct answer is - 15.625
Explanation:
The half-life is the time it takes to decay half value or quantity of its initial quantity or value. It is given that,
The half-life of K-42 = 12.4 hours
The initial value of K-42 = 500 gm
After 62 hours the value of k-42 = ?
The number of half cycle = 62*1/12.4 = 5,
therefore it will have 5 half-lives
initial - 500
1st half life - 500/2 = 250
2nd - half life - 250/2 = 125
3rd half life - 125/2 = 62.5
4th half life - 62/2 = 31.25
fifth half life - 31.25/2 = 15.625
Thus, the correct answer is =15.625g
15.625 g of a 500 gram sample of potassium-42 with 12.4 hours of Half -life is left after 62 hours.
The half-life is the time after which half of the initial quantity of a radioactive substance is left .
Given here,
The half-life of K-42 = 12.4 hours
The initial quantity of K-42 = 500 gm
The final quantity after 62 hours = ?
The number of half cycle
[tex]\bold { = 62\times \dfrac 1{12.4} = 5,}[/tex]
Thus the number of half life cycles in 62 hours is 5,
After 1 half life [tex]\bold {= \dfrac {500}{2} = 250}[/tex]
After 2 half lives [tex]\bold {= \dfrac {250}{2} = 125}[/tex]
After 3 half lives [tex]\bold {= \dfrac {125}{2} = 62.5}[/tex]
After 4 half lives [tex]\bold {= \dfrac {62.5}{2} = 31.25}\\[/tex]
After 5 half lives [tex]\bold {= \dfrac {31.25}{2} = 12.625}\\[/tex]
15.625 g of a 500 gram sample of potassium-42 with 12.4 hours of Half -life is left after 62 hours.
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13 Elements within the same main groups (1-8)
are cation often ligand in complex ?
Answer:
To name a coordination compound, no matter whether the complex ion is the cation or the anion.
Explanation:
A block of metal has a volume of 14.0 in' and weighs 5.28 lb.
Part A
What is its density in grams per cubic centimeter?
Express the density in grams per cubic centimeter to three significant figures.
Explanation:
1inch = 2.54cm
1inch^3= 2.54³ = 16.38cm^3
13.4cm^3 = 13.4*16.38 = 219.5867 cm^3
1lb = 454g
5.14lb = 5.14*454g/lb = 2,333.6g
density g/cm^3 = 2,333.6/219.6cm^3
density = 10.6265938 g/cm^3
to three significant figures:
density = 10.6g/cm^3
11. Which of the following is a characteristic of all chemical changes? *
Answer:
i think A
Explanation:
definitely not c
a new substance is not always formed, like when you burn a piece of paper.
i am stuck between a and b :(
How many grams of copper (II) fluoride (CuF2) are needed to make 9.66 liters of a 2.73 M
solution?
Mass of Copper (II) fluoride needed = = 2677.87 g
Further explanationGiven
9.66 L of 2.73 M Copper (II) fluoride
Required
Mass
Solution
Molarity : mol solutes per liters of solution
M = n : V
n = mol solute
V = volume of solution
mol Copper (II) fluoride :
= M x V
= 2.73 M x 9.66 L
= 26.3718 mol
Mass Copper (II) fluoride :
= mol x MW CuF₂
= 26.3718 mol x 101,543 g/mol
= 2677.87 g
Which best describes the three general groups of elements?
Answer:
D
Explanation:
2. Put the following steps of scientific inquiry in order, please number 1‐3.
_______Make an observation
_______ Ask a question
_______ Form a hypothesis
What type of charge do ionic compounds have
Answer:neutral
Explanation:ionic compounds is between a cation and anion so they cancel eachother out
