Aluminium is produced by electrolysis of a molten mixture.

How many grams of NO can be made from 50 g of NH3? 4 NH3 + 5 O2 → 6 H2O + 4 NO

Evgen [1.6K]

0.08 because 4mol NO divided by 50g gives you your answer

4 0

2 years ago

The Group 1B metals, Cu, Ag, and Au, are called coinage metals. What chemical properties make them especially suitable for makin

notka56 [123]

Answer:

Explanation:

Coinage metals -

The group 11 elements , i.e. , Copper , Silver , Gold are called the coinage metals .

These metals are quite soft and can be easily molded to form coins . This property is called as the property of malleability .

And due its soft nature , it can easily be used to make jewelry ,

Since , making jewels require the metal to be soft and flexible ,

Hence , gold is suited the best for this .

4 0

3 years ago

Oxygen gas is collected over water. The total pressure (the O2 pressure the water vapor pressure) is 736 torr. The temperature o

tigry1 [53]

Answer:

The value of the partial pressure of the oxygen $P_{O_{2} }$ = 690 torr

Explanation:

Total pressure of the mixture of gases = 736 torr

The partial pressure of water vapor = 46 torr

From the law of pressure we know that

Total pressure = The partial pressure of water vapor + The partial pressure of oxygen $O_{2}$

Put the values of pressures in above equation we get,

⇒ 736 = 46 + $P_{O_{2} }$

⇒ $P_{O_{2} }$ = 736 - 46

⇒ $P_{O_{2} }$ = 690 torr

This is the value of the partial pressure of the oxygen.

3 0

3 years ago

The neutrons within the nucleus of an atom have what charge?

vivado [14]

Answer:

the answer is D.

Explanation:

your welcome :)

4 0

2 years ago

During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.

Travka [436]

Answer: The mass of HCl present in 500 mL of acid solution is 36.5 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL$

Putting values in above equation, we get:

$1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

$2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g$

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

4 0

3 years ago