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3 years ago

What volume of a 0.33-M C12H22O11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M?

Chemistry

1 answer:

Vsevolod [243]3 years ago

7 0

The dilution formula can be used to find the volume needed

c1v1 = c2v2

Where c1 is concentration and v1 is volume of the concentrated solution

And c2 is concentration and v2 is volume of the diluted solution to be prepared

c1 - 0.33 M

c2 - 0.025 M

v2 - 25 mL

Substituting these values in the equation

0.33 M x v1 = 0.025 M x 25 mL

v1 = 1.89 mL

Therefore 1.89 mL of the 0.33 M solution needs to be diluted up to 25 mL to make a 0.025 M solution

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Which gas will effuse at the rate closest At a particular pressure and temperature, nitrogen gas effuses at the rate of 79mLs. U

Contact [7]

Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}$ ..........(1)

where,

$R_1$ = rate of effusion of nitrogen gas = $79mL/s$

$R_2$ = rate of effusion of sulfur dioxide gas = ?

$M_1$ = molar mass of nitrogen gas = 28 g/mole

$M_2$ = molar mass of sulfur dioxide gas = 64 g/mole

Now put all the given values in the above formula 1, we get:

$(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}$

$R_2=52mL/s$

Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.

4 0

3 years ago

In a certain city, electricity costs $0.15 per kW·h. What is the annual cost for electricity to power a lamp-post for 6.00 hours

Vanyuwa [196]

(a) Power of bulb is 100 W, converting this into kW.

$1 W=\frac{1}{1000}kW$

Thus,

$100 W =\frac{100}{1000}kW=0.1 kW$

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

$t=6\times 365=2190 h$

Electricity used will depend on power and number of hours as follows:

$E=P\times t=0.1 kW\times 2190 h=219 kW.h$

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 219 kW.h will be:

$Cost=\$ (219\times 0.15)=\$ 32.85$

Therefore, annual cost of incandescent light bulb is $\$ 32.85$

(b) Power of bulb is 25 W, converting this into kW.

$1 W=\frac{1}{1000}kW$

Thus,

$100 W =\frac{25}{1000}kW=0.025 kW$

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

$t=6\times 365=2190 h$

Electricity used will depend on power and number of hours as follows:

$E=P\times t=0.025 kW\times 2190 h=54.75 kW.h$

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 54.75 kW.h will be:

$Cost=\$ (54.75\times 0.15)=\$ 8.21$

Therefore, annual cost of fluorescent bulb is $\$ 8.21$ .

7 0

3 years ago

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose

Slav-nsk [51]

Answer:

The minimum mass of octane that could be left over is 43.0 grams

Explanation:

Step 1: Data given

Mass of octane = 73.0 grams

Mass of oxygen = 105.0 grams

Molar mass octane = 114.23 g/mol

Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate the number of moles

Moles = mass / molar mass

Moles octane = 73.0 grams / 114.23 g/mol

Moles octane = 0.639 moles

Moles O2 = 105.0 grams / 32.0 g/mol

Moles O2 = 3.28 moles

Step 4: Calculate the limiting reactant

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

O2 is the limiting reactant. It will completely be consumed. (3.28 moles). There will react 3.28 / 12.5 = 0.2624 moles. There will remain 0.639 - 0.2624 = 0.3766 moles octane

Step 5: Calculate mass octane remaining

Mass octane = moles * molar mass

Mass octane = 0.3766 moles * 114.23 g/mol

Mass octane = 43.0 grams

The minimum mass of octane that could be left over is 43.0 grams

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3 years ago

If you took a cross-section of soil from your backyard, what would be the correct order of soil layers from top to bottom? paren

AlladinOne [14]

humus, topsoil, subsoil, parent rock

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3 years ago

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Mercury-197 has a half-life of 3 days. Starting

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Answer:

2.34 gms left

Explanation:

3 weeks = 21 days = 7 half lives

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1 year ago