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Nikitich [7]
3 years ago
15

. A student had a 37.8g sample of compound X. The percent of nitrogen in this compound is 17.2%. How many grams of nitrogen are

present in this sample?
Chemistry
1 answer:
ArbitrLikvidat [17]3 years ago
5 0

Answer:

6.50 grams of the compound X is nitrogen

Explanation:

Step 1: Data given

Mass of compound X = 37.8 grams

Compound contains 17.2 % N

Step 2: Calculate mass of nitrogen

Mass nitrogen = % Nitrogen * total mass

Mass nitrogen = 17.2 % * 37.8 grams = 0.172 * 37.8 grams

Mass nitrogen = 6.50 grams

6.50 grams of the compound X is nitrogen

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An atom of argon has a radius rar = 88 pm and an average speed in the gas phase at 25°C of 172 m/s.
Rudik [331]

Answer:

1.2* 10³ rNe.

Explanation:

Given speed of neon=350 m/s

Un-certainity in speed= (0.01/100) *350 =0.035 m/s

As per heisenberg uncertainity principle

Δx*mΔv ≥\frac{h}{4\pi }

4π

h

..................(1)

mass of neon atom =\frac{20*10^{-3} }{6.22*10^{-23} } =3.35*10^{-26} kg

6.22∗10

−23

20∗10

−3

=3.35∗10

−26

kg

substituating the values in eq. (1)

Δx =4.49*10^{-8}10

−8

m

In terms of rNe i.e 38 pm= 38*10^{-12}10

−12

Δx=\frac{4.49*10^{-8} }{38*10^{-12} }

38∗10

−12

4.49∗10

−8

=0.118*10^{4}10

4

* (rNe)

=1.18*10³ rN

= 1.2* 10³ rNe.

Explanation:

This is the answer

7 0
2 years ago
Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is exce
adoni [48]

Answer:

The answer is "10.84 g".

Explanation:

The formula for calculating the number for moles:

\text{Number of moles }= \frac{\ Mass}{ \ molar \ mass }

In the given acid nitric:

Owing to the nitric acid mass = 75.9 g

Nitric acid molar weight= 63\  \frac{g}{mol}

If they put values above the formula, they receive:

\text{moles in nitric acid} = \frac{75.9}{63}

                               =1.204 \ mol

In the given chemical equation:

3 NO_2 \ (g) + H_2O \ (l) \longroghtarrow 2 HNO_3 \ (aq) + NO\ (g)

In this reaction, 2 mols of nitric acid are produced by 1 mole of water.

So, 1.204 moles of nitric acid will be produced:

= \frac{1}{2} \times 1.204

=0.602\ \ \text{mol of water}.

We are now using Equation 1 in determining the quantity of water:

Water moles = 0.602\  mol

Water weight molar = 18.02 \ \frac{g}{mol}

\to 0.602 = \frac{\text{mass of mols}}{ 18.02}\\\\\to \text{mass of mols} = 0.602 \times 18.02\\\\

                         =10.84 \ g

8 0
3 years ago
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