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3 years ago

Cobalt III hydroxide and nitric acid react according to the following balanced equation:

1 answer:

6 0

The mass of cobalt (III) needed is
m = 5.2 L (0.42 mol/L) ( 93 g/mol)
m = 97.65 g

The volume of nitric acid needed is
V = 5.2 L (0.42 mol/L) (3 mol / 1 mol) (1000 mL/1.6 mol)
V = 1968.75 mL

The moles of water produced is
n = 5.2 L (0.42 mol/L) (3 mol / 1 mol)
n = 3.15 moles

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Why did the big bang not produce a significant proportion of elements heavier than helium?

Travka [436]

The big bang did not produce a significant proportion of elements heavier than helium because the temperatures and densities present in the early universe were not sufficient to support the fusion of heavier elements.

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1 year ago

How much heat energy must be absorbed to completely melt 35.0 grams of H2O(s) at 0°C?

saul85 [17]

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3 years ago

The molarity of a solution prepared by dissolving 17.0 g of hydrochloric acid (HCl (aq)) in 133 mL of water is ___ M (2 decimal

Lera25 [3.4K]

The molarity is 3.50 M/L.

Explanation:

Molarity is found to know the amount of the solute ions present in the given volume of solution. So we determine it using the ratio of the moles of solute to the volume of the solution.

As here the weight of the solute which is hydrochloric acid is given as 17g, so we need to find the moles of it. This can be done by dividing the given amount of HCl with the molecular weight of HCl.

Molecular weight of HCl = 1.01 + 35.45 = 36.46 g/mol

So the number of moles of HCl present here will be

$No.of moles = \frac{17}{36.46} =0.466$

So, 0.466 moles of HCl is present in the solution.

$Molarity = \frac{No.of moles of solute}{Volume of solution} = \frac{0.466}{133*10^{-3} }$

$Molarity = 3.50 M/L$

Thus, the molarity is 3.50 M/L.

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4 years ago

Complete the balanced molecular chemical equation for the reaction below.

Alenkasestr [34]

Answer:

Following are the complete balance of the given equation:

Explanation:

Given equation:

$Cu_2SO_4(aq) + Li_3PO_4(aq)$

$Cu_2So4\ (aq)+Li_3Po_4 \(aq) \longrightarrow Cu_3(Po_4)\ (aq)+Li_2So_4 \ (aq)$

After Balancing the equation:

$3 Cu_2So4\ (aq)+ 2 Li_3Po_4\ (aq) \longrightarrow 2 Cu_3(Po_4)\ (aq)+ 3Li_2So_4 \ (aq)$

In the above equation, when the 3 mol Copper sulfate reacts with 2 mol lithium phosphate , it will produce 2 mol Copper phosphate and 3 mol Lithium sulfate .

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3 years ago