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Marysya12 [62]
2 years ago
14

What is the mass of 19.5g

Chemistry
1 answer:
ratelena [41]2 years ago
3 0

Answer:

I think 0.195 kg, if not then the question is not complete

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A 45ml of a 4M solution of CaBr2 contains how many grams of CaBr2?
natta225 [31]

The required mass of calcium bromide is 35.98 grams.

<h3>What is molarity?</h3>

Molarity is any solution is define as the number of moles of solute present in per liter of solution as;

M = n/V, where

  • M = molarity = 4M
  • V = volume = 45mL = 0.045L

Moles will be calculated by using the above equation as:

n = (4)(0.045) = 0.18 mole

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

  • W = given mass
  • M = molar mass

Mass of CaBr₂ = (0.18mol)(199.89g/mol) = 35.98g

Hence required mass of CaBr₂ is 35.98 grams.

To know more about molarity, visit the below link:
brainly.com/question/22283918

#SPJ1

6 0
2 years ago
A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
Is that the right answer
julsineya [31]
Is what the right answer
8 0
2 years ago
Use the periodic table and example bond table to answer the question.
lyudmila [28]

In keeping with the general trends, K-Br will have the smallest bond energy. The bond energy refers to the energy that keeps the atoms in a bond together.

<h3>What is bond energy?</h3>

Bond energy is the energy that is required to hold atoms together in a bond. This energy must also be supplied when the atoms are to be separated.

We have the bond energies of each of the bons in the question, we have to note that the the smallest value of bond energy is Na-Br hence in keeping with the general trends, K-Br will have the smallest bond energy.

Learn more about bond energies: brainly.com/question/14842720?

7 0
1 year ago
Scientific explanations are not always based on empirical evidence <br><br> True or false?
tigry1 [53]
True some explanations are not always based on empirical evidence
4 0
3 years ago
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