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3 years ago

How many moles of sodium are present in 17 g of Na?

Chemistry

1 answer:

11Alexandr11 [23.1K]3 years ago

8 0

Answer:

1. 0.74mol

2. 0.42mol

3. 2.125mol

4. 0.301mol

5. 4.52 × 10^23 particles

Explanation:

Number of moles (n) in a substance can be found using the formula:

mole (n) = mass/molar mass

Using this formula, the following moles are calculated:

1. Molar of Na = 23g/mol

mole = 17/23

mole = 0.74mol

2. Molar mass of Na2SO4 = 23(2) + 32 + 16(4)

= 46 + 32 + 64

= 142g/mol

Mole = 60/142

mole = 0.42mol

3. Molar mass of CO2 = 12 + 16(2)

= 12 + 32

= 44g/mol

mole = 93.5/44

mole = 2.125mol

4. Molar mass of sodium nitrate (NaNO3) = 23 + 14 + 16(3)

= 23 + 14 + 48

= 85g/mol

mole = 25.6/85

mole = 0.301mol

5. Number of particles in one mole of a substance is 6.022 × 10^23 particles. Hence, in 0.75mol of calcium hydroxide (Ca(OH)2, there will be;

0.75mol × 6.02 × 10^23

= 4.515 × 10^23

= 4.52 × 10^23 particles

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Answer:

See figure 1

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3 years ago

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.CCl4(g) + (1/2)O2

Elodia [21]

Answer : The value of $K'_c$ for the reaction is, $1.9\times 10^{19}$

Explanation :

The given equilibrium reactions are:

(1) $CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g)$ $K_c=4.4\times 10^9$

(2) $2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$ $K'_c=?$

As we know that, when we are multiplying the equation (1) by 2 then we get the equation (2). Thus, the relation between the two equilibrium constant will be,

$K'_c=(K_c)^2$

So,

$K'_c=(4.4\times 10^9)^2$

$K'_c=1.9\times 10^{19}$

Therefore, the value of $K'_c$ for the reaction is, $1.9\times 10^{19}$

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4 years ago