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3 years ago

What is the net ionic equation for the following? :)

Chemistry

1 answer:

I am Lyosha [343]3 years ago

3 0

Answer:

1. Mg (s) + 2Na+(aq) → 2Na(s) + Mg²⁺(aq)

2. 2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)

Explanation:

The net ionic equation of a reaction express only the chemical species that are involved in the reaction:

1. Mg (s) + Na2CrO4 (aq) → 2Na + MgCrO4(aq)

The ionic equation:

Mg (s) + 2Na+(aq) + CrO4²⁻ (aq) → 2Na + Mg²⁺ + CrO4²⁻(aq)

Subtracting the ions that don't change:

2. 2K(s) + Cd(NO3)2(aq) → 2KNO3(aq) + Cd(s)

The ionic equation:

2K(s) + Cd²⁺(aq) + 2NO3⁻(aq) → 2K⁺(aq) + 2NO3⁻(aq) + Cd(s)

Subtracting the ions that don't change:

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Calculate the volume in milliliters of a iron(II) bromide solution that contains of iron(II) bromide . Round your answer to sign

miv72 [106K]

This is an incomplete question, here is a complete question.

Calculate the volume in milliliters of a 1.29 mol/L iron(II) bromide solution that contains 275 mmol of iron(II) bromide . Round your answer to significant 3 digits.

Answer : The volume of iron(II) bromide solution is, $2.13\times 10^2mL$

Explanation : Given,

Concentration of iron(II) bromide = 1.29 mo/L

Moles of iron(II) bromide = 275 mmol = 0.275 mol

conversion used : 1 mmol = 0.001 mol

Now we have to calculate the volume of iron(II) bromide.

$\text{Volume of iron(II) bromide}=\frac{Moles of iron(II) bromide}}{\text{Concentration of iron(II) bromide}}$

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

$\text{Volume of iron(II) bromide}=\frac{0.275mol}{1.29mol/L}=0.213L=2.13\times 10^2mL$

Thus, the volume of iron(II) bromide solution is, $2.13\times 10^2mL$

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3 years ago

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sladkih [1.3K]

Answer:

See explanation

Explanation:

20 = B

21 = C

22 = D

23 = B

24 = B

25 = C

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