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3 years ago

Answer both 8 and 9 will give u brain list with explanation as well so don’t just answer

Chemistry

1 answer:

goldfiish [28.3K]3 years ago

6 0

Answer:

8. the answer is B.

9. the answer is A.

Explanation:

Hello!

8. In this case, by bearing to mind that the limiting reactant is always completely consumed and the excess one remain as a leftover at the end of the reaction, we can also infer that as all the limiting reactant is consumed, it must determine the maximum amount of product as the excess reactant will hypothetically produce a greater mass than expected; thus, the answer to this question is B.

9. In this case, since the mole ratio of oxygen to water is 1:2, the following proportional factor is used to calculate the produced mass of water:

$3molO_2*\frac{2molH_2O}{1molO_2}=6molH_2O$

Thus, the answer is this case is A.

Best regards!

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Define an ion in chemistry

REY [17]

Ion is an atom or molecule with a net electric charge due to loss or gain of one or more electrons.

4 0

3 years ago

How many moles of water can be produced with 4.3 moles of H2 and 5.6 moles of O2? Which reactant is limiting? How many moles of

xeze [42]

Answer:

Hydrogen H₂ will be the limiting reagent.

The excess reactant that will be left after the reaction is 3.45 moles.

4.3 moles of water can be produced.

Explanation:

The balanced reation is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

H₂: 2 moles
O₂: 1 mole
H₂O: 2 moles

To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?

$moles of H_{2} =\frac{5.6 moles of O_{2} *2 mole of H_{2} }{1 mole of O_{2}}$

moles of H₂= 11.2 moles

But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, hydrogen H₂ will be the limiting reagent and oxygen O₂ will be the excess reagent.

Then you can apply the following rules of three:

If by reaction stoichiometry 2 moles of H₂ react with 1 mole of O₂, 4.3 moles of H₂ will react with how many moles of O₂?

$moles of O_{2} =\frac{1 mole of O_{2} *4.3 mole of H_{2} }{2 mole of O_{2}}$

moles of O₂= 2.15 moles

The excess reactant that will be left after the reaction can be calculated as:

5.6 moles - 2.15 moles= 3.45 moles

The excess reactant that will be left after the reaction is 3.45 moles.

If by reaction stoichiometry 2 moles of H₂ produce 2 moles of H₂O, 4.3 moles of H₂ produce how many moles of H₂O?

$moles of H_{2}O =\frac{2 moles of H_{2}O *4.3 mole of H_{2} }{2 mole of H_{2}}$

moles of H₂O= 4.3 moles

4.3 moles of water can be produced.

8 0

3 years ago

How many milliliters of 2.19 M H2SO4 are required to react with 4.75 g of solid containing 21.6 wt% Ba(NO3)2 if the reaction is

Setler [38]

Answer:

1.7927 mL

Explanation:

The mass of solid taken = 4.75 g

This solid contains 21.6 wt% $Ba(NO_3)_2$ , thus,

Mass of $Ba(NO_3)_2$ = $\frac {21.6}{100}\times 4.75\ g$ = 1.026 g

Molar mass of $Ba(NO_3)_2$ = 261.337 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.026\ g}{261.337\ g/mol}$

$Moles= 0.003926\ mol$

Considering the reaction as:

$Ba(NO_3)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3$

1 moles of $Ba(NO_3)_2$ react with 1 mole of $H_2SO_4$

Thus,

0.003926 mole of $Ba(NO_3)_2$ react with 0.003926 mole of $H_2SO_4$

Moles of $H_2SO_4$ = 0.003926 mole

Also, considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Molarity = 2.19 M

So,

$2.19=\frac{0.003926}{Volume\ of\ the\ solution(L)}$

Volume = 0.0017927 L

Also, 1 L = 1000 mL

So, volume = 1.7927 mL

5 0

3 years ago

Al2O3 would be typed in as Al2O3 Al2(SO4)3 would be typed in as Al2(SO4)3.

N76 [4]

Answer:

Explanation:

Hello,

We are required to write the chemical formula of the following compounds

1. Calcium nitride = Ca₃N₂

2. Magnesium Phosphide = Mg₃P₂

3. Rubidium Chromate = Rb₂CrO₄

4. Aluminium nitrate = Al(NO₃)₃

5. Ammonium Arsenide = (NH₄)₃As

6. Nickel(ii) nitrite = Ni(NO₂)₂

7. Copper(i)sulfate = Cu₂SO₄

8. Iron(iii)nitrate = Fe(NO₃)₃

9. Manganese(ii)nitrate = Mn(NO₃)₂

The left hand side are the common names when the right hand side are the chemical formula.

4 0

3 years ago

Calculate to three significant digits the density of chlorine pentafluoride gas at exactly and exactly . You can assume chlorine

slega [8]

Answer:

Density is 6.16g/L

Explanation:

... at exactly -15°C and exactly 1atm...

Using general gas law:

PV = nRT

We can find density (Ratio of mass and volume) in an ideal gas as follows:

P/RT = n/V

To convert moles to grams we need to multiply the moles with Molar Weight, MW:

n*MW = m

n = m/MW

P/RT = m/V*MW

P*MW/RT = m/V

Where P is pressure: 1atm;

MW of chlorine pentafluoride: 130.445g/mol

R is gas constant: 0.082atmL/molK

And T is absolute temperature: -15°C+273.15 = 258.15K

Replacing:

P*MW/RT = m/V

1atm*130.445g/mol / 0.082atmL/molK*258.15K = m/V

6.16g/L = m/V

<h3>Density of the gas is 6.16g/L</h3>

7 0

3 years ago