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castortr0y [4]
3 years ago
8

Write a balanced net ionic equation for the following reaction: SrCl2(aq) + H2CO3(aq) → SrCO3(s) + HCl (aq)

Chemistry
1 answer:
max2010maxim [7]3 years ago
5 0
  We will balance the equation in the following order: metals, amethals, carbon, hydrogen and oxygen (the most common order).
  The metal present in the equation is Sr, which is already balanced (there are 1 on each side of the equation).
  The amethal present in the equation is Cl. There is 2 Cl in the left side and only one in the right side. So, we will multiply the quantity of the molecule that contains Cl by 2. Doing this, we'll obtain:

SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}
 
  Looking at the equation, we can see that it is now fully balanced. Hence, a balanced equation of the reaction is:

SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}
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True water is wet because when something is wet in this case it's water it has water on it at a molecular level. Water molecules are bonded on top of each other so it's wet.

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4 0
3 years ago
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DaniilM [7]

Answer:

625 mL

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V₁) = 250 mL

Molarity of stock solution (M₁) = 5 M

Molarity of diluted solution (M₂) = 2 M

Volume of diluted solution (V₂) =?

The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

5 × 250 = 2 × V₂

1250 = 2 × V₂

Divide both side by 2

V₂ = 1250 / 2

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8 0
3 years ago
he heat of fusion of tetrahydrofuran is . Calculate the change in entropy when of tetrahydrofuran melts at . Be sure your answer
lana [24]

Answer:

\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}

Explanation:

Hello.

In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:

\Delta S=\frac{n*\Delta H}{T}

Whereas n accounts for the moles which are computed below:

n=5.9g*\frac{1mol}{72g} =0.082mol

Thus, the entropy turns out:

\Delta S=\frac{0.0819mol*8.5 kJ/mol}{(-108.5+273.15)K}\\\\\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}

Best regards.

3 0
3 years ago
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