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3 years ago

To what volume should you dilute 50.0 ml of 12 m hno3 solution to obtain a 0.100 m hno3 solution?

Chemistry

1 answer:

bearhunter [10]3 years ago

4 0

Answer:

The answer is "6L"

Explanation:

Formula:

$\bold{C_1 \times V_1 = C_2 \times V_2 }\\\\V_2= \frac{C_1\times V_1}{C_2}$

Values:

$\to C_1= 12 \ m\\\to V_1= 50 \ ml\\\to C_2= 0.100 \ m\\\\\\V_2= \frac{12 \times 50 }{0.100}$

$= \frac{12 \times 50 }{0.100}\\\\= \frac{12 \times 50 \times 1000}{100}\\\\= \frac{600 \times 1000}{100}\\\\= 600 \times 10\\\\=6000 \ ml\\= 6 \ L$

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On combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical

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Answer:

0.2 is conpound Co2 STP.

Explanation:

on combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical formula of compound.?

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2 years ago

You have collected a tissue specimen that you would like to preserve by freeze drying. To ensure the integrity of the specimen,

Reika [66]

Answer:

The maximum pressure is 612.2 Pa

Explanation:

The pressure of the ice (P1) = 624 Pa

The temperature of the ice = 273.16 K

The maximum temperature the specimen = - 5 oC

= -5 + 273 = 268K

The maximum Pressure the freeze drying can be will be (P2) = ?

Using Pressure law, which shows the relationship between pressure and temperature.

P1 / T1 = P2 / T2

P2 T1 = P1 T2

P2 = P1 T2 / T1

P2 = 624 × 268 / 273.16

P2 = 612.2 Pa

The maximum pressure at which drying can be carried out is 612.2 Pa

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3 years ago

A sample of hydrated copper (II) sulfate (CuSO4•nH2O) is heated to 150°C and produces 103.74 g anhydrous copper (II) sulfate and

wariber [46]

Answer:

Explanation:

Firstly, we convert what we have to percentage compositions.

There are two parts in the molecule, the sulphate part and the water part.

The percentage compositions is as follows:

Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%

The water part = 100 - 64 = 36%

Now, we divide the percentages by the molar masses.

For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol

For the H2O = 2(1) + 16 = 18g/mol

Now we divide the percentages by these masses

Sulphate = 64/160 = 0.4

Water = 36/18 = 2

The ratio is thus 0.4:2 = 1:5

Hence, there are 5 water molecules.

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3 years ago

35. In the collision theory, a collision that leads to the formation of products is called an

FinnZ [79.3K]

Answer:

It's Effective Collision.

Explanation:

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3 years ago

Chlorine is used to disinfect swimming pools. the accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 g of chl

frez [133]

The accepted concentration of chlorine is 1.00 ppm that is 1 gram of chlorine per million of water.

The volume of water is $2.29\times 10^{4} gal$ .

Since, 1 gal= 3785.41 mL

Thus, $2.29\times 10^{4} gal=2.29\times 10^{4}\times 3785.41 mL=8.66\times 10^{7}mL$

Density of water is 1 g/mL thus, mass of water will be $8.66\times 10^{7}g$ .

Since, 1 grams of chlorine → $10^{6}$ grams of water.

1 g of water → $10^{-6}$ g of chlorine and,

$8.66\times 10^{7}g$ of water →86.6 g of chlorine

Since, the solution is 9% chlorine by mass, the volume of solution will be:

$V=\frac{100}{9}\times 86.6 mL=9.62\times 10^{2} mL$

Thus, volume of chlorine solution is 9.62\times 10^{2} mL.

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3 years ago