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nalin [4]
3 years ago
12

What is the molarity of a solution that contains 0.082 mol KI in a 2.03 L solution?

Chemistry
1 answer:
astraxan [27]3 years ago
4 0

Answer:

molarity of the KI solution = 0.04 mol/L

Explanation:

Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution.  

The law that we can applied to calculate the M is:  

M = n / V

n - number of moles

V- volume of the solution (liters)

Then insert in the equation the values from the question;  

M = 0.082 mol  /  2.03 L = 0.04 mol/L

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By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask
tangare [24]

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = V_1=11.00 mL

Concentration of stock solution = M_1=0.823 M

Volume of stock solution after dilution = V_2=50.00 mL

Concentration of stock solution after dilution = M_2=?

M_1V_1=M_2V_2 ( dilution )

M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L

0.823 Molar is the molarity of the solution.

6 0
3 years ago
Why do the planets keep orbiting the sun
dsp73
There's no reason, it's just because the sun has a gravitational pull
3 0
3 years ago
Solar system question
shtirl [24]

Answer:

yes

Explanation:

is that is the alteration of a planet on its orbit around the sun depends upon the mass of the Sun and the inverse square of the planets distance from the Sun. As a planet moves further away in its orbit around the Sun, the gravitational force exerted by the sun on the planet decreases

8 0
3 years ago
: If a 250. mL sample of the above buffer solution initially has 0.0800 mol H2C6H5O7 - and 0.0600 mol HC6H5O7 2- , what would be
Jet001 [13]

Answer: New concentration of HC_{6}H_{5}O^{2-}_{7} is 0.23 M.

Explanation:

The given data is as follows.

     Moles of HC_{6}H_{5}O^{2-}_{7} = 0.06 mol

     Moles of H_{2}C_{6}H_{5}O_{7} = 0.08 mol

Therefore, moles of OH^{-} added are as follows.

    Moles of OH^{-} = 0.125 \times \frac{25}{1000}

                          = 0.003125 mol

Now, new moles of HC_{6}H_{5}O^{2-}_{7} = 0.06 + 0.003125

                     = 0.063125

Therefore, new concentration of HC_{6}H_{5}O^{2-}_{7} will be calculated as follows.

       Concentration = \frac{0.063125}{0.275}

                               = 0.23 M

Thus, we can conclude that new concentration of HC_{6}H_{5}O^{2-}_{7} is 0.23 M.

4 0
3 years ago
Calculate the volume in ml of 0.200 M na2co3 needed to produce 2.00 g of caco3 there is an excess of cacl2
UkoKoshka [18]

Answer:

V=100mL

Explanation:

Hello.

In this case, since the chemical reaction is:

Na_2CO_3+CaCl_2\rightarrow  CaCO_3+2NaCl

We next compute the moles of sodium carbonate from the 2.00 grams of calcium carbonate via their 1:1 mole ratio in the chemical reaction:

n_{Na_2CO_3}=2.00gCaCO_3*\frac{1molCaCO_3}{100.09gCaCO_3}*\frac{1molNa_2CO_3}{1molCaCO_3}  \\\\n_{Na_2CO_3}=0.0200molNa_2CO_3

Thus, by knowing the molarity, we compute the volume:

M=\frac{n}{V}\\ \\V=\frac{n}{M}=\frac{0.0200mol}{0.200mol/L}\\  \\V=0.100L*\frac{1000mL}{1L}\\ \\V=100mL

Best regards.

8 0
3 years ago
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