How many atoms are in 3.5 mol he?

Chemistry

1 answer:

hichkok12 [17]2 years ago

3 0

Answer:

2.11×10^24

Explanation:

You times the Avogadro number by 3.5

6.023 times 10^23.

then times it by 3.5.

I got 2.11 times 10 ^24.

I hope this helps. Have a lovely day.

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What is the percentage composition when 10g of magnesium combines with 4g of nitrogen?

Ivanshal [37]

The % composition when 10g of magnesium combine with 4g of nitrogen is 71.43% magnesium and 28.57 % nitrogen

calculation

% composition = mass of an element / total mass x100
mass of magnesium = 10 g
mass of nitrogen = 4g

calculate the total mass used

= 10g of Magnesium + 4 g of nitrogen = 14 grams

% composition for magnesium is therefore = 10/14 x100 = 71.43 %

% composition for nitrogen is therefore = 4 /14 x100 = 28.57 %

7 0

3 years ago

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$