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3 years ago

How many equivalents are in 0.60 mole of Mg2+

1 answer:

5 0

Equivalents express the combining power of an element for example what mass of an element can combine with or displace one gram of hydrogen or eight grams of oxygen.

Since Mg is divalent, one gram of Mg2+ can combine with 2 grams of hydrogen

Therefore, 0.6 mole of Mg2+ can combine with 0.6 x 2 = 1.2 moles of hydrogen

Hence there are 1.2 equivalents in 0.60 mole of Mg2+

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A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o

statuscvo [17]

Answer: The value of $K_{eq}$ is 0.044

Explanation:

We are given:

Initial moles of methane = $4.10\times 10^{-2}mol=0.0410moles$

Initial moles of carbon tetrachloride = $6.51\times 10^{-2}mol=0.0651moles$

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of methane = $\frac{0.0410}{1.00}=0.0410M$

Concentration of carbon tetrachloride = $\frac{0.0651}{1.00}=0.0651M$

The given chemical equation follows:

$CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)$

Initial: 0.0410 0.0651

At eqllm: 0.0410-x 0.0651-x 2x

We are given:

Equilibrium concentration of carbon tetrachloride = $6.02\times 10^{-2}M=0.0602M$

Evaluating the value of 'x', we get:

$\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M$

Now, equilibrium concentration of methane = $0.0410-x=[0.0410-0.0049]=0.0361M$

Equilibrium concentration of $CH_2Cl_2=2x=[2\times 0.0049]=0.0098M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044$

Hence, the value of $K_{eq}$ is 0.044

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Which of the following is not the same as 0.00684 kg? 0.0684 dkg,6.84 g, 684 cg, 6,840 mg,

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The Answer would be C.) 684 cg

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I need help FAST ASAP

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In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.

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2. C⁴⁺ , O²⁻ , CO₂ ---> carbon dioxide

3. Al³+ , Cl⁻ , AlCl₃ ----> aluminum chloride

4. Ca²⁺ , Cl⁻ , CaCl₂ ---> calcium chloride

5. Li⁺ , Br⁻ , LiBr ---> lithium bromide

6. Mg³⁺ , O²⁻ , Mg₂O₃ ----> magnesium oxide

7. K⁺ , I⁻ , KI ---> potassium iodide

8. H⁺ , Cl⁻ , HCl --> hydrogen chloride

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10. Na⁺ , Br⁻ , NaBr ---> sodium bromide

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