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2 years ago

How many molecules are contained in 1.55 moles of methane, CH 4 ?

Chemistry

1 answer:

Korolek [52]2 years ago

5 0

Avogadro's law states that in one mole of a substance, there are $6.022 \times 10^{23}$ molecules.

This means that in 1.55 moles, there are $1.55(6.022 \times 10^{23})=\boxed{9.33 \times 10^{23} \text{( to 3 sf)}}$

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Potassium iodide reacts with lead(II) nitrate in this precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Wha

andrew11 [14]

Answer:

a. 174 mL

Explanation:

Let's consider the following reaction.

2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)

We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:

0.1550 L × 0.112 mol/L = 0.0174 mol

The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:

2 × 0.0174 mol = 0.0348 mol

The volume of a 0.200 M KI solution that contains 0.0348 moles is:

0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL

5 0

2 years ago

Did I get these correctly?

Luda [366]

Those are both correct! great job, keep up the good work (-:

4 0

2 years ago

What term is used for the electrons in the outermost shell

son4ous [18]

Answer:

valence electrons

Explanation:

4 0

2 years ago

Read 2 more answers

Whuch expression is equal to the number of grams in 2.43 kg

schepotkina [342]

2.43*1000.

There are 1000 grams in a kilogram

7 0

3 years ago

One of relatively few reactions that takes place directly between two solids at room temperature is In this equation, the in Ba(

yuradex [85]

Answer:

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate.

Explanation:

$Ba(OH)_2.8H_2O(s)+NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+8H_2O(l)+NH_3(g)$

The balance chemical equation is :

$Ba(OH)_2.8H_2O(s)+2NH_4SCN(s)\rightarrow Ba(SCN)_2(s)+10H_2O(l)+2NH_3(g)$

Mass of barium hydroxide octahydrate = 6.5 g

Moles of barium hydroxide octahydrate = $\frac{6.5 g}{315 g/mol}=0.020635 mol$

According to reaction, 2 moles of ammonium thiocyanate reacts with1 mole of barium hydroxide octahydrate. The 0.020635 moles of barium hydroxide octahydrate will react with:

$\frac{2}{1}\times 0.020635 mol=0.04127 mol$

Mass of 0.04127 moles of ammonium thiocyanate;

$0.04127 mol\times 76 g/mol=3.136 g\approx 3.14 g$

3.14 grams of ammonium thiocyanate must be used to react completely with 6.5 g barium hydroxide octahydrate

3 0

3 years ago