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3 years ago

How many molecules are in 41.8 g H2O?

Chemistry

2 answers:

Sergio [31]3 years ago

8 0

Answer:

7.63 × 10²³ molecules

Explanation:

First, convert grams to moles using the molar mass of water (32.988 g/mol).

41.8 g ÷ 32.988 g/mol = 1.267 mol

Next, convert moles to molecules using Avogadro's number (6.022 × 10²³).

1.267 mol × 6.022 × 10²³ molecules/mol = 7.63 × 10²³ molecules

photoshop1234 [79]3 years ago

7 0

Answer:

7.63 × 10²³ molecules

Explanation:

hey

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At a certain temperature, 0.800 mol SO 3 is placed in a 1.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equil

Elanso [62]

Answer : The value of equilibrium constant (K) is, 0.004

Explanation :

First we have to calculate the concentration of $SO_3\text{ and }O_2$

$\text{Concentration of }SO_3=\frac{\text{Moles of }SO_3}{\text{Volume of solution}}=\frac{0.800mol}{1.50L}=1.2M$

and,

$\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{0.150mol}{1.50L}=0.1M$

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

$2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)$

Initial conc. 1.2 0 0

At eqm. (1.2-2x) 2x x

As we are given:

Concentration of $O_2$ at equilibrium = x = 0.1 M

The expression for equilibrium constant is:

$K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}$

Now put all the given values in this expression, we get:

$K_c=\frac{(2x)^2\times (x)}{(1.2-2x)^2}$

$K_c=\frac{(2\times 0.1)^2\times (0.1)}{(1.2-2\times 0.1)^2}$

$K_c=0.004$

Thus, the value of equilibrium constant (K) is, 0.004

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4 years ago

Concentrated hydrochloric acid is an aqueous solution that is 34.70 % HCl. The density of the solution is 1.19 g/mL. What mass o

mixas84 [53]

The mass of HCl that is contained in the solution is 147 g HCl

Why?

To find the mass of HCl we have to apply what is called a conversion factor. In a conversion factor we put the units we don't want at the bottom, and the ones we want at the top.

For this question, we want to go from liters of solution to mass of HCl, and the conversion factor is laid out as follows:

$0.356Lsolution*\frac{1000mL}{1L}*\frac{1.19 g solution}{1 mL solution}*\frac{34.70 g HCl}{100 g solution}=147 g HCl$

Have a nice day!

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