This image shows the rock cycle.Which event most likely occurs at point V?

Hydrazine, N2H4, may react with oxygen to form nitrogen gas and water.

Sindrei [870]

Answer:

The percent yield of the reaction is 35 %

Explanation:

In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.

Let's verify the moles that were used in the reaction.

2.05 g . 1mol/ 32 g = 0.0640 mol

In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.

Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).

1atm . 0.550L = n . 0.082 . 295K

(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles

Percent yield of reaction = (Real yield / Theoretical yield) . 100

(0.0225 / 0.0640) . 100 = 35%

3 0

3 years ago

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URGENT:

Ad libitum [116K]

Answer:

The elements that are commonly positive ions are metals. But there are a few gases that can become positively charged by losing electrons.

8 0

3 years ago

How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Lady_Fox [76]

Answer:

$\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}$

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

Na₂ = 22.9897693 * 2= 45.9795386 g/mol
O₃ = 15.999 * 3= 47.997 g/mol
Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

$\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

We are converting 57.3 grams to moles, so we multiply by this value.

$57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

Flip the ratio so the units of grams of sodium carbonate cancel.

$57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}$

$57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }$

$\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3$

$0.5406295944 \ mol \ Na_2CO_3$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

$0.541 \ mol \ Na_2CO_3$

There are approximately 0.541 moles of sodium carbonate in 57.3 grams.

6 0

3 years ago

If a gas is moved from a large container to a small container but its temperature and number of moles remain the same, what woul

bazaltina [42]

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

P2 = P1V1/V2

The correct answer is the first option. Pressure would increase. This can be seen from the equation above where V2 is indirectly proportional to P2.

8 0

3 years ago

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For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

Charra [1.4K]

Answer:

heres the link of the answer

boldniwally.com

3 0

3 years ago