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Delvig [45]
3 years ago
7

Through electrolysis, a student collects 6.6 g of hydrogen gas and 52.4 g of oxygen gas. The reaction was allowed to proceed to

completion. What was the mass of water initially present?
Chemistry
1 answer:
Kaylis [27]3 years ago
3 0

Answer: 59 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

2H_2O\rightarrow 2H_2+O_2

Given: mass of hydrogen = 6.6 g

mass of oxygen = 52.4 g

Mass of products = Mass of hydrogen + mass of oxygen = 6.6 +52.4 = 59 g grams

Thus mass or reactant = mass of water

Mass of reactants = mass of products = 59 g

Thus the mass of water initially present was 59 g.

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How does the rock cycle and plate tectonics interfere with geochemists trying to determine the age of the Earth? Chose the best
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4 0
2 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
2 years ago
how many grams of molecular oxygen (O2) is produced from 13.8 grams of calcium chlorate (Ca(ClO3)2)in the following chemical rea
olga2289 [7]

Answer:

6.72 g

Explanation:

Given data:

Mass of calcium chlorate = 13.8 g

Mass of oxygen produced = ?

Solution:

Chemical equation:

Ca(ClO₃)₂        →      CaCl₂ + 3O₂

Number of moles of calcium chlorate:

Number of moles = mass / molar mass

Number of moles = 13.8 g/ 206.98 g/mol

Number of moles = 0.07 mol

Now we will compare the moles of oxygen and  calcium chlorate.

                 Ca(ClO₃)₂         :            O₂

                      1                   :              3

                    0.07               :            3×0.07=0.21 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.21 mol × 32 g/mol

Mass = 6.72 g

3 0
2 years ago
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