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0.15 gm of metallic oxide was dissolved in 100 ml of 0.1 N H2SO4 and 25.8 ml of 0.095N NaOH were used to neutralise the remainin

Mila [183]

Answer:

Explanation:

25.8 ml of .095 N NaOH is needed to neutralise the remaining acid

equivalent of NaOH used = 25.8 x .095 / 1000 = .002451 gm equivalent .

acid remaining = .002451 gm equivalent .

acid initially taken = 100 ml of .1 N / 1000 = . 01 gm equivalent

acid reacted with metal = .01 -.002451 = .007549 gm equivalent

This must have reacted with same gram equivalent of metal oxide

.007549 gm equivalent = .15 gm of metal oxide

1 gm equivalent = 19.87 gm

equivalent weight of metal = 19.87 - equivalent weight of oxygen

= 19.87 - 8 = 11.87 .

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7 0

2 years ago

What is the osmotic pressure of a solution prepared from 13.7 g of the electrolyte HCl and enough water to make 0.500 L of solut

kherson [118]

Answer:

P = 17.9618 atm

Explanation:

The osmotic pressure can be calculated and treated as if we are talking about an ideal gas, and it's expression is the same:

pV = nRT

However the difference, is that instead of using moles, it use concentration so:

p = nRT/V ----> but M = n/V so

p = MRT

We have the temperature of 18 °C (K = 18+273.15 = 291.15 K) the value of R = 0.08206 L atm / K mol, so we need to calculate the concentration, and we have the mass of HCl, so we use the molar mass of HCl which is 36.45 g/mol:

n = 13.7/36.45 = 0.3759 moles

M = 0.3759/0.5 = 0.7518 M

Now that we have the concentration, let's solve for the osmotic pressure:

p = 0.7518 * 0.08206 * 291.15

p = 17.9618 atm

3 0

2 years ago

N2 and H2 are mixed in 14:3 mass ratio. After certain time ammonia was found to be 40% by mol. The mole fraction of N2 at that t

son4ous [18]

Answer : The mole fraction of nitrogen will be 0.4615.

Explanation : When nitrogen ( $N_{2}$ )and hydrogen ( $H_{2}$ )are mixed, the mole ratio becomes 1 : 1.5,

Now we know that ( $H_{2}$ ) is acting as a limiting agent.

So at the time of when 0.4 moles of ( $NH_{3}$ ) is been formed it requires 0.4 moles of ( $N_{2}$ ) and 3.4 moles of ( $H_{2}$ )

So, we find the the remaining ( $N_{2}$ ) will be 0.6 and

( $H_{2}$ ) will be 0.3 mole present in mixture.

So, the mole fraction of ( $N_{2}$ ) becomes = 0.6 / (0.6 + 0.4 + 0.3) Which becomes = 0.4615

8 0

2 years ago

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If 6.00 g of the unknown compound contained 0.200 mol of C and 0.400 mol of H, how many moles of oxygen, O, were in the sample?

Arada [10]

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

Answer:

0.2 moles O

8 0

2 years ago

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As ice transforms into steam, A)new molecules are formed. B)the types of atoms are changed. C)energy is absorbed by the molecule

ra1l [238]

This is an example of a substance changing state, a physical change, the molecules are changed, but the atoms themselves do not change, just their arrangement, and the mass of the molecules is the same. Therefor energy is absorbed by the molecules, as energy is required to change the state or physicality of a molecule structure.Hope this helps. Let me know if you need additional help!

8 0

2 years ago

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