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Maslowich
3 years ago
6

A chemist prepares a solution of barium chloride by measuring out of barium chloride into a volumetric flask and filling the fla

sk to the mark with water. Calculate the concentration in of the chemist's barium chloride solution. Round your answer to significant digits.
Chemistry
1 answer:
mezya [45]3 years ago
8 0

The given question is incomplete. The complete question is:

A chemist prepares a solution of barium chloride by measuring out 110 g of barium chloride into a 440 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mole per liter of the chemist's barium chloride solution. Round your answer to 3 significant digits.

Answer: Concentration of the chemist's barium chloride solution is 1.20 mol/L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

Molarity=\frac{n\times }{V_s}

where,

n = moles of solute

V_s = volume of solution in L

moles of BaCl_2(solute) = \frac{\text {given mass}}{\text {Molar Mass}}=\frac{110g}{208g/mol}=0.529mol

Now put all the given values in the formula of molality, we get

Molality=\frac{0.529\times 1000}{440ml}=1.20mole/L

Therefore, the molarity of solution is 1.20 mol/L

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How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction
nekit [7.7K]

Answer : The volume of NaOH required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?

Putting values in above equation, we get:

2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL

Hence, the volume of NaOH required to neutralize is, 340 mL

4 0
3 years ago
The American Heart Association recommends to eat no more than 2,301mg of sodium per day. Convert the mass of
castortr0y [4]

The mass of 2,301 grams of sodium in ounces is 0.0811757609 ounces.

The amount of sodium recommended by American heart association is 2301 mg. This limit should not be crossed in a day.

We have to convert Mass of sodium from mg to ounces.

We know,

1 ounce = 28.3459 grams.

We also know,

1 gram = 1000 milligrams.

So,

28.3459 grams = 28.3459 x 1000 milligrams.

28.3459 grams = 28345.9 milligrams.

1 ounce = 28345.9 mg.

1 mg = 1/28345.9 ounces

Weight of sodium 2301mg in ounces,

2301mg = 2301/28345.9 ounces

Dividing till last significant figure,

2301 mg = 0.0811757609 ounces.

Mass of sodium in ounces is 0.0811757609.

To know more about Unit conversion, visit,

brainly.com/question/97386

#SPJ9

4 0
1 year ago
At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide:
gayaneshka [121]

<u>Answer:</u> The moles of bromine gas at equilibrium is 0.324 moles.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}        .......(1)

Calculating the initial moles of hydrogen and bromine gas:

  • <u>For hydrogen gas:</u>

Moles of hydrogen gas = 0.682 mol

Volume of solution = 2.00 L

Putting values in equation 1, we get:

\text{Molarity of solution}=\frac{0.682mol}{2.00L}=0.341M

  • <u>For bromine gas:</u>

Moles of bromine gas = 0.440 mol

Volume of solution = 2.00 L

Putting values in equation 1, we get:

\text{Molarity of solution}=\frac{0.440mol}{2.00L}=0.220M

Now, calculating the molarity of hydrogen gas at equilibrium by using equation 1:

Equilibrium moles of hydrogen gas = 0.566 mol

Volume of solution = 2.00 L

Putting values in equation 1, we get:

\text{Molarity of solution}=\frac{0.566mol}{2.00L}=0.283M

Change in concentration of hydrogen gas = 0.341 - 0.283 = 0.058 M

This change will be same for bromine gas.

Equilibrium concentration of bromine gas = (\text{Initial concentration}-\text{Change in concentration})=0.220-0.058=0.162M

Now, calculating the moles of bromine gas at equilibrium by using equation 1:

Molarity of bromine gas = 0.162 M

Volume of solution = 2.00 L

Putting values in equation 1, we get:

0.162M=\frac{\text{Moles of bromine gas}}{2.00L}\\\\\text{Moles of bromine gas}=(0.162mol/L\times 2.00L)=0.324mol

Hence, the moles of bromine gas at equilibrium is 0.324 moles.

5 0
3 years ago
Most of the ultraviolet radiation reaching the surface of the earth is UV-A radiation, which has a wavelength range of 315 nm to
posledela

Answer:

6.1 ×10^-19 J

Explanation:

From E= hc/λ

h= planks constant = 6.6×10^-34 Js

c= speed of light = 3×10^8 ms^1

λ= wavelength = 325 nm

E= 6.6 × 10^-34 × 3×10^8/325 × 10^-9

E= 0.061 × 10^ -17 J

E= 6.1 ×10^-19 J

4 0
3 years ago
In the reaction below 25.00 g of MnC1, is reacted with 100.0 g of PbO2, excess KCl, and excess
HACTEHA [7]

Answer:

31.44 g KMnO4

Explanation:

M(MnCl2) = M(Mn) + 2M(Cl) = 54.9 + 2*35.5 =125.9 g/mol

25.00 g /125.9 g/mol =0.1990 mol MnCl2

M(PbO2) = M(Pb) + 2M(O2) = 207.2 +2*16.0 =239.2 g/mol

100g/139.2 g/mol = 0.7184 mol PbO2

              2 KCl + 2 MnCl2 + 5 PbO2 + 4 HCl → 2 KMnO4 + 5 PbC12 + 2 H2O

from reaction          2 mol          5 mol

given                    0.1990mol    0.7184 mol

for 2 mol MnCl2         ----- 5 mol PbO2

for 0.1990 mol MnCl2  ---- x mol PbO2

x = (0.1990 *5)/2 = 0.4975 mol PbO2

So, for 0.1990 MnCl2 we need 0.4975 mol PbO2, but we have 0.7184 mol PbO2. That means that we have excess of PbO2, and we are going to use for further calculation 0.1990 mol MnCl2

                2 KCl + 2 MnCl2 + 5 PbO2+ 4 HCl → 2 KMnO4 + 5 PbC12 + 2 H2O

from reaction        2 mol                                       2 mol

given                       0.1990 mol

gotten                                                                  0.1990 mol

We got 0.1990 mol KMnO4.

M(HMnO4) = M(K) + M(Mn) +4M(O) = 158.0 g/mol

m(KMnO4) = 0.1990 mol*158.0 g/mol = 31.44 g

7 0
3 years ago
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