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3 years ago

How many grams of hydrogen are contained in 2.00 mol of C6H7N

Chemistry

1 answer:

Setler [38]3 years ago

3 0

Answer:- 14.0 moles of hydrogen present in 2.00 moles of $[tex]C_6H_7N$ .

Solution:- We have been given with 2.00 moles of $C_6H_7N$ and asked to calculate the grams of hydrogen present in it. It's a two step conversion problem. In first step we convert the moles of the compound to moles of hydrogen as one mol of the compound contains 7 moles of hydrogen. In next step the moles are converted to grams on multiplying the moles by atomic mass of H. The calculations are shown as:

$2.00molC_6H_7N(\frac{7molH}{1molC_6H_7N})(\frac{1.0gH}{1molH})$

= 14.0 g H

So, there are 14.0 g of hydrogen in 2.00 moles of $C_6H_7N$ .

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Nitrogen effuses through a pinhole 1.7 times as fast as another gaseous element under the same conditions. Estimate the other el

marysya [2.9K]

Answer:

80.92, Krypton

Explanation:

What is effusion?

• It is a process where gas escapes through a pinhole (a very small hole) into a region of low pressure or vacuum

Graham's law of effusion of gas

• states that at a given constant temperature and pressure, the rate of effusion of gases is inversely proportional to the square root of their molar masses

$\boxed{ \frac{Rate_1}{Rate_2} = \sqrt{ \frac{M_2}{M_1} } }$

Calculations

Nitrogen exist as N₂ at room temperature, thus its molar mass is 2(14)= 28.

Let the rate and molar mass of unknown gas be Rate₂ and M₂ respectively.

Since N₂ effuses 1.7 times as fast as the unknown gas,

Rate₁= 1.7(Rate₂)

$\frac{Rate_1}{Rate_2} = 1.7$

$1. 7 = \sqrt{ \frac{M_2}{28} }$

Square both sides:

$2.89 = \frac{M_2}{28}$

Multiply both sides by 28:

2.89(28)= M₂

M₂= 80.92

Identity of gas

The molar mass of 80.92 lies between Bromine and Krypton. However since Bromine exist as Br₂, the value of it's molar mass would be 159.8 instead. Hence, Bromine is eliminated.

If the gas is a diatomic element, the atomic weight is 80.92 ÷2= 40.46. Thus, we are now considering if Argon could be its identity. However, Argon is a noble gas and will not exist as a diatomic element. Argon is therefore eliminated too.

Thus based on the above reasoning, its probable identity is Krypton.

7 0

3 years ago

Which shows that friction is undesirable?

Nataly_w [17]

C. Walking on wet floors shows that Friction is undesirable

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2 years ago

Which numerical setup can be used to calculate the atomic mass of the element bromine?

Gekata [30.6K]

Answer:

From the numerical steps highlighted under explanation, the average atomic mass of bromine is 79.91 u

Explanation:

The steps to be taken will involve;

1) Find the number of isotopes of bromine.

2) Identify the atomic mass and relative abundance of each of the isotopes.

3) Multiply the atomic mass of each of the isotopes by their corresponding values relative abundance value.

4) Add the value in step 3 above to get the average atomic mass of bromine.

Now;

Bromine has 2 isotopes namely;

Isotope 1: Atomic mass = 78.92amu and a relative abundance of 50.69%.

Isotope 2: Atomic mass = 80.92amu and a relative abundance of 49.31%.

Using step 3 above, we have;

(78.92 × 50.69%)

And (80.92 × 49.31%)

Using step 4 above, we have;

(78.92 × 50.69%) + (80.92 × 49.31%) ≈ 79.91 u

6 0

3 years ago

How many formula units are present in 25.0 mol of sodium chloride

bonufazy [111]

The way to work out formula units is to do the amount of moles (in this case 25) times by 6.022 x 10^23. so, the answer is 1.5055 x 10^25 formula units. hope this helps! (by the way ^ represents ‘to the power of’ because i cant get the smaller symbols on here)

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3 years ago

A compound is found to contain 64.80 % carbon, 13.62 % hydrogen, and 21.58 % oxygen by weight. What is the empirical formula for

insens350 [35]

Answer:

Empirical formula is C₄H₁₀O

Explanation:

Values for C, H and O are determined as centesimal composition.

64.80 g of C in 100g of compound

13.62g of H in 100 g of compound

21.58 g of O in 100 g of compound.

We convert the mass to moles:

64.80 g . 1mol/ 12g = 5.4 moles of C

13.62 g . 1 mol /1g = 13.62 moles of H

21.58 g . 1 mol/16g = 1.35 moles of O

We pick the lowest value and we divide:

5.4 moles of C / 1.35 = 4 C

13.62 moles of H / 1.35 = 10 H

1.35 moles of O / 1.35 = 1 O

Empirical formula is C₄H₁₀O, it can be the diethyl ether.

We confirm, the excersise is well done.

Molar mass = 74g/mol

74 g of compound we have (12 . 4)g of C

In 100 g of compound we may have (100 . 48) / 74 = 64.8 g

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3 years ago