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Nitric oxide is made from the oxidation of ammonia. What mass of nitric oxide can be made from the reaction of 8.00 g NH 3 with

kvv77 [185]

Answer: The mass of nitric oxide is 12.72 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For ammonia:

Given mass of ammonia = 8.00 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonia}=\frac{8.00g}{17g/mol}=0.47mol$

For oxygen gas:

Given mass of oxygen gas = 17.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{17g}{32g/mol}=0.53mol$

The given chemical reaction follows:

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$

By Stoichiometry of the reaction:

5 moles of oxygen gas reacts with 4 moles of ammonia

So, 0.53 moles of oxygen gas will react with = $\frac{4}{5}\times 0.53=0.424mol$ of ammonia

As, given amount of ammonia is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

5 moles of oxygen gas produces 4 moles of NO

So, 0.53 moles of oxygen gas will produce = $\frac{4}{5}\times 0.53=0.424moles$ of NO

Now, calculating the mass of NO from equation 1, we get:

Molar mass of NO = 30 g/mol

Moles of NO = 0.424 moles

Putting values in equation 1, we get:

$0.424mol=\frac{\text{Mass of NO}}{30g/mol}\\\\\text{Mass of NO}=(0.424mol\times 30g/mol)=12.72g$

Hence, the mass of nitric oxide is 12.72 grams

6 0

4 years ago

The numeral 4.21 has three significant fire two known figures d e figure numbers are always significant

Kamila [148]

The significant figures are always:

Different from zero except there are only zeros before the point.

You can round them to the previous significant.

In scientific notation, you have one figure point two more figures.

Examples:

You have 4.21

All different from zero and only two decimals.

Those are all significant figures.

if you have 000.231555

You will shorten this to two significant figures.

Before the point, you only have zeros, so you need to keep only one of them to say its less than one.

After the point, you have a lot of figures, but you need to round this to two.

Because you have a one before the three, you'll keep the three. If you have a five or bigger number, you round it.

In this case, you'll have 0.23 with two significant figures.

7 0

4 years ago

Which would have more volume? A mass of mercury or the same mass of bromine. The density of mercury is 13.5g/ml. The density of

deff fn [24]

Explanation:

Given that,

The density of mercury is 13.5 g/mL

The density of Bromine is 3.12 g/cm³

It is mentioned that Mercury and bromine have the same mass. Let d₁,d₂ are the density of Mercury and Bromine. V₁ and V₂ are their volumes. So,

$\text{density}=\dfrac{\text{mass}}{\text{volume}}$

$\text{mass}=\text{density}\times \text{volume}$

Since, mass is same.

So,

$d_1V_1=d_2V_2\\\\\dfrac{d_1}{d_2}=\dfrac{V_2}{V_1}\\\\\dfrac{13.5}{3.12}=\dfrac{V_2}{V_1}\\\\4.32=\dfrac{V_2}{V_1}\\\\V_2=4.32\times V_1$

Hence, the volume of bromine is more than that of mercury. It is 4.32 times of the density of mercury.

3 0

3 years ago

If the aluminum block is initially at 25 ∘C∘C, what is the final temperature of the block after the evaporation of the alcohol?

Effectus [21]

Answer:

Final temperature of aluminum block = 12.1°C

Note: The question is not complete. The complete question is given below:

If the aluminum block is initially at 25 ∘C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘C. Heat of vaporization of the alcohol at 25 ∘C is 45.4 kJ/mol Suppose that 1.12 g of rubbing alcohol (C3H8O) evaporates from a 73.0 g aluminum block.

Explanation:

Heat lost by aluminum block = heat required for vaporization of alcohol

Heat required to vaporize ethanol, H = mass of alcohol * heat of vaporization of alcohol

Mass of alcohol = 1.12 g; molar mass of rubbing alcohol = 60 g/mol

Heat of vaporization = (45.4 kJ/mol)/ 60 g/mol = 0.75666 kJ/g

H = 1.12 g × 0.7566 kJ/g

H = 0.8474 kJ = 847.4 J

Heat lost by aluminum block, Q = -(mass × specific heat capacity × temperature change)

Mass of aluminum block = 73.0 g

Specific heat capacity of aluminum = 0.900 J/g

Temperature change = (Final temperature, T - 25)

Q = -73.0 g × 0.900 J/g × (T - 25)

Q = -65.7 J × (T - 25°C)

Since, Heat lost by aluminum block = heat for vaporization of alcohol

-65.7 J × (T - 25°C) = 847.4 J

T - 25 = 847.4/-65.7

T - 25 = -12.9

T = -12.9 + 25

T = 12.1°C

5 0

3 years ago

Which refrigerant is known to destroy ozone?

Degger [83]

Answer:

tetrafluoroethane

Explanation:

Its an HCFC, which are the refrigerants that destroy the ozone layer.

8 0

3 years ago

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