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Answer:

1.807×10and at the top of ten put -24

Answer:

MP hippopotamus

Explanation: I'm in the fifth grade so yeah I really don't know this so I'm just going to say some random stuff and white cheddar mac and cheese warm TV perfect back MD 11443 to 2885 eleven 12:20 to 11 132 0.24 answer

Answer:

False

Explanation:

Water vapour exists in gas phase!

Answer:

Even in a dry desert environment, the water cycle is taking place. The winds here blow up snow from the land and put it into the atmosphere, which is part of the water cycle. And the sun helps out, too, causing sublimation to occur, which causes snow to evaporate directly into water vapor gas

Explanation:

2.04 mol CBr₄

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Organic

Atomic Structure

Step 1: Define

675 g CBr₄

Step 2: Identify Conversions

Molar Mass of C - 12.01 g/mol

Molar Mass of Br - 79.90 g/mol

Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol

Step 3: Convert

[tex]\displaystyle 675 \ g \ CBr_4(\frac{1 \ mol \ CBr_4}{331.61 \ g \ CBr_4}) = 2.03552 \ mol \ CBr_4[/tex]

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

2.03552 mol CBr₄ ≈ 2.04 mol CBr₄

Answer:

[tex]\boxed {\boxed {\sf About \ 2.04 \ moles \ of \ CBr_4}}[/tex]

Explanation:

1. Define Formula

The compound is carbon tetrabromide. The tetra indicates 4 atoms of bromine.

2. Find Molar Mass

There are 2 elements in this compound: carbon and bromine. Use the Periodic Table to find the molar masses of these elements.

The molar mass is based on the number of atoms in the compound. The compound has 1 atom of carbon and 4 atoms of bromine.

CBr₄= 1(12.011 g/mol) + 4(79.90 g/mol)

= 12.011 g/mol+319.60 g/mol = 331.611 g/mol

3. Convert Grams to Moles

We want to convert 675 grams to moles. We should use the molar mass as a fraction.

[tex]\frac{331.611 \ g \ CBr_4}{1 \ mol \ CBr_4 }[/tex]

Multiply by the given number of grams.

[tex]675 \ g \ CBr_4 *\frac{331.611 \ g \ CBr_4}{1 \ mol \ CBr_4 }[/tex]

Flip the fraction so the grams of CBr₄ will cancel.

[tex]675 \ g \ CBr_4 *\frac{1 \ mol \ CBr_4 }{331.611 \ g \ CBr_4}[/tex]= [tex]675 \ *\frac{1 \ mol \ CBr_4 }{331.611 \ }[/tex]

[tex]\frac{675 \ mol \ CBr_4 }{331.611 \ }[/tex] = [tex]2.03551752 \ mol \ CBr_4[/tex]

4.Round

The original measurement, 675 grams has 3 significant figures (6,7 and 5), so our answer must have the same.

For the answer we found, 3 sig figs is the hundredth place.

[tex]2.03551752 \ mol \ CBr_4[/tex]

The 5 in the thousandth place tells us to round the 3 to a 4.

[tex]\approx 2.04 \ mol \ CBr_4[/tex]

There are about 2.04 moles of carbon tetrabromide in 675 grams.

Answer:

- P in PBr3 is +3.

- N in N2 is 0.

- As in H3AsO4 is +5.

Explanation:

Hello!

In this case, since the determination of the oxidation states is performed by using the well-known charge balances, we can proceed as shown below:

- P in PBr3: Here, bromide ions have an oxidation state of -1, so we follow:

[tex]P^xBr_3^-\\\\x-3=0\\\\x=+3[/tex]

Thus, the oxidation state is +3.

- N in N2: Here, since nitrogen is bonding with nitrogen and it is neutral, we infer its oxidation state is 0.

- As in H3AsO4: Here, oxygen is -2 and hydrogen +1, so we follow:

[tex]H_3^+As^xO_4^{-2}\\\\3+x-8=0\\\\x=8-3\\\\x=5[/tex]

Thus, the oxidation state is +5.

Best regards!

Answer:

C3H8 + 5O2 ‐‐‐‐‐‐‐‐> 4H2O + 3CO2

Answer:

A weak acid

Explanation:

Im not sure if its right. But lets see someone else do it.

Answer:

B. A strong acid.

An electrolyte with a pH less than 4 is a strong acid, that with pH range of 4 - 7 is a weak acid, that with 7 - 10 is a weak base, that with pH range of 10 - 14 is a strong base.

Solution :

lt is given that in 18 mL of water their are [tex]6.022\times 10^{23}[/tex] water molecules.

We know, that 1 molecule of water contains 2 atoms of hydrogen.

Hydrogen atom in 18 mL water is,

[tex]2\times 6.022\times 10^{23} = 12.044 \times 10^{23}[/tex] .

So, number of hydrogen atoms in 1 L = 1000 mL are :

[tex]N = \dfrac{1000}{18}\times 12.044\times 10^{23}\\\\N = 6.69 \times 10^{25}\ atoms[/tex]

Hence, this is the required solution.

The given question is incomplete. the complete question is : If 5.15 g[tex]Fe(NO_3)_3[/tex] is dissolved in enough water to make exactly 323 ml of solution, what is the molar concentartion of nitrate ion.

Answer:  The molar concentartion of nitrate ion is 0.195.

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Given : 5.15 g of [tex]Fe(NO_3)_3[/tex] is dissolved

Volume of solution = 323 ml

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n= moles of solute

[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{5.15g}{242g/mol}=0.021moles[/tex]  

[tex]V_s[/tex] = volume of solution = 323 ml

[tex]Molarity=\frac{0.021\times 1000}{323}=0.065M[/tex]

As 1 M of [tex]Fe(NO_3)_3[/tex] gives 3 M of [tex]NO_3^-[/tex] ions.

Thus 0.065 M of [tex]Fe(NO_3)_3[/tex] gives  = [tex]\frac{3}{1}\times 0.065=0.195M[/tex] of [tex]NO_3^-[/tex] ions.

The molar concentartion of nitrate ion is 0.195.

Answer:

Slide a magnet through the mixture. When you take the magnet off, the iron filings are going to be attracted to the magnet and the sand will stay in the bowl

Explanation:

Answer:

Density can be used to separate the substances that make up a mixture, because each substance in a mixture has its own density.  For example, if a mixture of sand and oil is placed in water, the sand will sink to the bottom of the container. The sand is more dense than the water.

Explanation:

Please mark me as brainy if this helps

Answer:

62.9%

Explanation:

Step 1: Given data

Step 2: Calculate the mass lost of water

We will use the following expression.

mH₂O = mNa₂CO₃.xH₂O - mNa₂CO₃

mH₂O = 71.5 g - 26.5 g = 45.0 g

Step 3: Calculate the percent of water in the hydrate sodium carbonate

We will use the following expression.

%H₂O = mH₂O / mNa₂CO₃.xH₂O × 100%

%H₂O = 45.0 g / 71.5 g × 100%

%H₂O = 62.9%

Answer:

100

Explanation:

Answer:

The First Battle of Panipat was fought between the invading forces of Babur and the Lodi Empire, which took place on 21 April 1526 in North India. It marked the beginning of the Mughal Empire. This was one of the earliest battles involving gunpowderfirearms and field artillery.

For you to write an essay on  Panipat first bring into view

Then write a brief story using this format.

In writing this essay, you must consider the structure and content of the essay or short note

For structure kick off with

In conclusion, for the content of the short note you can consider this,

Panipat fought its first battle on 21 April 1526 in the North of India, This battle was fought between the invading powers of Babur and the Lodi Empire.

Read more on History

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Answer:false

Explanation:

Answer:

A the first one

Explanation:

Answer:

The Answer is fjords

Explanation:

Hope It Helps! Make Me Braninest

Answer:

[tex]SF_5[/tex]

Explanation:

Hello!

In this case, when determining empirical formulas by knowing the by-mass percent, we can first assume we have 25.24 g of sulfur and 74.76 g of fluorine, so we compute the moles:

[tex]n_S=25.24g*\frac{1mol}{32.07g} =0.787mol\\\\n_F=74.76g*\frac{1mol}{19.00g}= 3.935mol[/tex]

Now, we divide the moles of both S and F by the fewest moles, in this case, those of S in order to determine the subscripts in the empirical formula:

[tex]S=\frac{0.787}{0.797}=1 \\\\F=\frac{3.935}{0.797}=5[/tex]

Therefore, the empirical formula is:

[tex]SF_5[/tex]

Best regards!

Answer: The possible molecular formula will be [tex]CO_2[/tex]

Explanation:

Mass of C= 27.3 g

Mass of O = 72.7 g

Step 1 : convert given masses into moles.

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{27.3g}{12g/mole}=2.275moles[/tex]

Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{72.7g}{16g/mole}=4.544moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{2.275}{2.275}=1[/tex]

For O =[tex]\frac{4.544}{2.275}=2[/tex]

The ratio of C : O = 1: 2

Hence the empirical formula is [tex]CO_2[/tex]

The possible molecular formula will be=[tex]n\times CO_2[/tex]

Answer:

Tellurium

Explanation:

Edg 2021

Answer:

Structurally, plant and animal cells are very similar because they are both eukaryotic cells. They both contain membrane-bound organelles such as the nucleus, mitochondria, endoplasmic reticulum, golgi apparatus, lysosomes, and peroxisomes. Both also contain similar membranes, cytosol, and cytoskeletal elements.

Explanation:

I hope it's help

Answer:

4.99*10²³ molecules of N₂O₄ are in 76.3 g of N₂O₄

Explanation:

Avogadro's Number is the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number represents a quantity without an associated physical dimension. Avogadro's number applies to any substance.

You know that the molar mass of N₂O₄ is 92.02 g/mol, and you have 76.3 g. Then you can apply the following rule of three: 92.02 grams are present in 1 mole of the compound, 76.3 grams in how many moles are they?

[tex]amount of moles= \frac{76.3 grams*1 mole}{92.02 grams}[/tex]

amount of moles= 0.83 moles

Then, you can apply another rule of three: if by definition of Avogadro's number 1 mole of the compound has 6.023*10²³ molecules, 0.83 moles of the compound, how many molecules will it have?

[tex]amount of molecules= \frac{0.83 moles*6.023*10^{23}molecules }{1 mole}[/tex]

amount of molecules= 4.99*10²³

4.99*10²³ molecules of N₂O₄ are in 76.3 g of N₂O₄

Answer:  The precipitate formed is [tex]BaSO_4[/tex]

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid or precipitated form are represented by (s) after their chemical formulas.  

A double displacement reaction in which one of the product is formed as a solid is called as precipitation reaction.  

The balanced chemical equation is:

[tex]K_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+2KCl(aq)[/tex]

An acid and a base which differ from each other by a proton are said to form a conjugate pair. The conjugate base of HSO₃⁻ is SO₃²⁻. The correct option is C.

The entity produced when an acid gives up a proton has a tendency to regain the proton and hence it behaves as a base. It is called the conjugate base of the parent acid.

The entity produced when a base accepts a proton has a tendency to donate the proton and hence behaves as an acid. It is called the conjugate acid of the parent base.

Sulfur dioxide dissolved in water gives sulfurous acid H₂SO₃. The conjugate base of this material contains one proton less i.e, HSO₃⁻ and the conjugate base of the substance bisulfite is SO₃²⁻which is sulfite ion.

Thus the correct option is C.

To know more about conjugate base, visit;

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The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- =  0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of  MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since  

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution =  5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

Explanation:

c realsed from the galaxie

Answer:

C

Explanation:

A P E X

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