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Evgesh-ka [11]
3 years ago
5

Solutions of mercury (I) nitrate and potassium bromide are mixed

Chemistry
1 answer:
Afina-wow [57]3 years ago
7 0
With that informatio you can:

1) Write the chemical equation
2) Balance the chemical equation
3) State the molar ratios
4) Predict if precipitation occurs.

I will do all four, for you:

1) Chemical equation:

mercury(I) nitrate  potassium bromide    mercury(I) bromide  potassium nitrate
<span>Hg2(NO3)2             +        KBr                   → Hg2Br2          +      KNO<span>3

2) Balanced chemical equation
</span></span>
<span>Hg2(NO3)2 + 2KBr → Hg2Br2 + 2KNO<span>3

3) Molar ratios or proportions:

1 mol </span></span><span>Hg2(NO3)2 : 2 mol KBr : 1 mol Hg2Br2 : 2 mol KNO<span>3

4) Prediction of precipitation.

You can use the solubility rules or a table of solubilities. I found in a table of solutiblities that mercury(I) bromide is insoluble and potassium bromide is soluble, Then you can predict that the precipitation of mercury(I) bromide will occur.


</span></span>
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The reaction for photosynthesis producing glucose sugar and oxygen gas is:
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<u><em>1.5 grams of glucose is produced from 2.20 g of CO₂.</em></u>

To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

So, in this case, the balanced reaction is:

6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:

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So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:

moles of CO_{2} =2.20 grams*\frac{1 mole}{44.01 grams}

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Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?

moles of C_{6} H_{12} O_{6} =\frac{0.05moles of CO_{2} *1 mole of C_{6} H_{12} O_{6}}{6moles of CO_{2}}

moles of C₆H₁₂O₆= 8.33*10⁻³

Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:

mass of glucose =8.33*10^{-3} moles*\frac{180.18 grams}{1 mole}

<em>mass of glucose= 1.5 grams</em>

Then, <u><em>1.5 grams of glucose is produced from 2.20 g of CO₂.</em></u>

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