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arsen [322]
3 years ago
7

Calculate the molality of a 15.0% by mass solution of MgCl 2 in H 2O. The density of this solution is 1.127 g/mL. Group of answe

r choices
Chemistry
1 answer:
saul85 [17]3 years ago
3 0

Answer:

1.86 m

Explanation:

Step 1: Calculate the mass of solute and solvent in 100 grams of solution

We have a 15.0% by mass solution of MgCl₂ (solute) in H₂O (water). In 100 g of solution, there are 15.0 g of solute and 100.0 g - 15.0 g = 85.0 g of solvent.

Step 2: Calculate the moles corresponding to 15.0 g of MgCl₂

The molar mass of MgCl₂ is 95.21 g/mol.

15.0g \times \frac{1mol}{95.21 g} = 0.158mol

Step 3: Convert the mass of water to kilograms

We will use the relationship 1 kg = 1,000 g.

85.0g \times \frac{1kg}{1,000g} = 0.0850kg

Step 4: Calculate the molality of the solution

m = \frac{0.158mol}{0.0850kg} = 1.86m

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Answer:

95.7 g CO to the nearest tenth.

Explanation:

2C + O2 ---> 2CO

Using relative atomic masses:

24 g C produces  2*12 + 2*16 g CO.

So 41 g produces  ( (2*12 + 2*16) * 41  ) / 24

= 95.7 g CO,

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3 years ago
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4 years ago
A sample of hydrogen gas at a pressure of 0.520 atm and a temperature of 26.2°C, occupies a volume of 15.4 liters. If the gas is
Oksanka [162]

Answer:

1.99 atm

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial pressure (P1) = 0.520 atm

Initial temperature (T1) = 26.2°C

Initial volume (V1) = 15.4L

Final temperature (T2) = constant = 26.2°C

Final volume (V2) = 4.02L

Final pressure (P2) =..?

Step 2:

Determination of the new pressure of the gas.

Since the temperature of the gas is constant, it means the gas is obeying Boyle's law. Thus, the new pressure of the gas can be obtained by applying the Boyle's law equation as shown below:

P1V1 = P2V2

0.520 x 15.4 = P2 x 4.02

Divide both side by 4.02

P2 = (0.520 x 15.4) / 4.02

P2 = 1.99 atm

Therefore, the new pressure of the gas is 1.99 atm

4 0
4 years ago
A substance stores ___________ as a result of its chemical bonds.
Maksim231197 [3]

Answer:

Chemical energy

3 0
3 years ago
Calculate the pH for each of the cases in the titration of 25.0 mL of 0.180 M pyridine, C 5 H 5 N ( aq ) with 0.180 M HBr ( aq )
Oxana [17]

Answer:

Explanations

Calculate the pH for each of the following cases in thetitration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 MHBr(aq):

a.Before and addition of HBr

b.After addition of 12.5ml HBr

c.After addition of 15ml HBr

d.After addition of 25ml HBr

e.After addition of 33ml HBr

SOLUTION ;;;

Kb of pyridine =1.5*10^-9

a)let the dissociation be x.so,

kb=x^2/(0.1-x)

or 1.5*10^-9=x^2/(0.1-x)

or x=1.225*10^-5

so,

[OH-]=1.225*10^-5

so,

pOH=-log([OH-])

so pH=9.088

b)now this will effectively behave as a buffer

pKb=8.82

so pOH=pKb+log(salt/acid)

=8.82+log((12.5*0.1)/(25*0.1-12.5*0.1))

=8.82

so pH=14-pOH

=5.18

c)again using the same equation as the above,

pOH=pKb+log(salt/acid)

=8.82+log((15*0.1)/(25*0.1-15*0.1))

=9

so pH=14-9

=5

d)now the base is completely neutralised.so,

concentration of the salt formed=0.1/2

=0.05 M

so,

pH=7-0.5pKb-0.5log(C)

=7-0.5*8.82-0.5*log(0.05)

=3.24

e)concentration of H+=(33*0.1-25*0.1)/(33+25)

=0.01379

so pH=-log(0.01379)

=1.86

4 0
4 years ago
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