Balance the chemichal equation:<br> NH3+CuO➡️Cu+N2+H2O

If I have 6 moles of a gas at a pressure of 5.8 atm and a volume of 12 liters, what

poizon [28]

The temperature of the system is 141.36K.

<h3><u>Explanation</u>:</h3>

According to the universal gas law, the pressure of a gas is directly proportional to the absolute temperature and number of moles of the gas and inversely proportional to the volume of the gas.

So in equation format,

$\frac{P\times V}{n\times T}$ =constant.

Or $\frac{P\times V}{n\times T}$ = R.

Here R is called the universal gas constant which has different values in different units.

Here the unit of volume is in litres and unit of pressure is in atmosphere.

So. Value of R comes as =0.082057 $\frac{L\times atm} {K\times mole}$ .

So, according to the question,

P is 5.8atm, V is 12 litres and n is 6 moles.

So, $5.8\times12$ = $6\times0.082057\times T$ .

Or, T = $\frac{5.8\times12} {6\times0.082057}$ .

Or, T = 141.36.

So,the temperature of the system is 141.36K

7 0

4 years ago

True or false during reactions or light is needed carbon dioxide is converted to glucose

Helen [10]

Answer:

True

Explanation:

7 0

3 years ago

If I have 22,9 moles of a gas at a pressure of 12.4 atm and a volume of 24 liters, what is the temperature?

Grace [21]

Answer:

160 K, using 2 SF,

Explanation:

Ideal Gas law, rearrange to solve for T (K), answer about 158.3 round to 2SF

PV = nRT

8 0

2 years ago

If you start with 227.8 grams of iron and 128 grams of oxygen to produce iron oxide, what is the limiting reagent?

agasfer [191]

The balanced chemical equation between iron and oxygen to produce iron (III) oxide is,

$4Fe(s) + 3O_{2}(g) ---> 2Fe_{2}O_{3}(s)$

Mass of Fe = 227.8 g

Moles of Fe = $227.8 g Fe * \frac{1 mol Fe}{55.85 g Fe} = 4.079 mol Fe$

Mass of oxygen = 128 g

Moles of $O_{2} = 128 g O_{2}*\frac{1 mol O_{2}}{32 g O_{2}}= 4mol O_{2}$

Calculating the limiting reactant: The reactant that produces the least amount of product will be the limiting reactant.

Mass of iron (III) oxide produced from Iron = $4.079 mol Fe * \frac{2 mol Fe_{2}O_{3}}{4 mol Fe} *\frac{159.69 g Fe_{2}O_{3}}{1 mol Fe_{2}O_{3}} = 325.7 g Fe_{2}O_{3}$

Mass of iron (III) oxide produced from oxygen= $4 mol O_{2}*\frac{2 molFe_{2}O_{3}}{3 mol O_{2}}*\frac{159.69 g Fe_{2}O_{3}}{1 mol Fe_{2}O_{3}} = 425.84 g Fe_{2}O_{3}$

Iron (Fe) produces the least amount of the product iron (III) oxide. So, Fe is the limiting reactant.

7 0

3 years ago

Whats the answers for brainpop skeletal system

choli [55]

Answer:

whats the question??? I am familiar with brainpop but if u tell me the question ill give u the asnwer

Explanation:

8 0

2 years ago

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Balance the chemichal equation: NH3+CuO➡️Cu+N2+H2O