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2 years ago

A student mixed together aqueous solutions of Y and Z. A white precipitate(solid)formed. which could not be Y and Z

1 answer:

maks197457 [2]2 years ago

4 0

(A)hydrochloric acid + silver nitrate
HCl(aq) + AgNO3(aq) -----> AgCl(s) +HNO3(aq)

(B)hydrochloric acid + sodium hydroxide
HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)

(C)calcium chloride + silver nitrate
CaCl2(aq) + AgNO3(aq) ----> AgCl(s) +Ca(NO3)2(aq)

(D)sodium chloride + silver nitrate
NaCl(aq) + AgNO3(aq) ----> AgCl(s) +NaNO32(aq)

AgCl is a white precipitate.
In (B) no precipitate was formed, so answer is B.

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A hot gas flowing through a pipeline can be considered as a:________

k0ka [10]

Answer:

B) irreversible process

Explanation:

The process given here is irreversible.

7 0

1 year ago

Which of the following molecules can form hydrogen bonds?

frosja888 [35]

Hydrogen bonding is a type of intermolecular forces of attraction in which hydrogen atom is bonded to one of the most electronegative atoms. This gives a partial positive charge to hydrogen atom and a partial negative charge to the electronegative atom involved in the bonding. The electronegative atoms that can form hydrogen bonding are fluorine (F), nitrogen (N), and oxygen (O).

Therefore the correct option is,

A) NH3

3 0

1 year ago

A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T

ladessa [460]

Answer:

$\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}$

Explanation:

1. Molar concentration

Let's call chloroform C and acetone A.

Molar concentration of C = Moles of C/Litres of solution

(a) Moles of C

Assume 0.187 mol of C.

That takes care of that.

(b) Litres of solution

Then we have 0.813 mol of A.

(i) Mass of each component

$\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}$

(ii) Volume of each component

$\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}$

(iii) Volume of solution

If there is no change of volume on mixing.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

$c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}$

2. Molal concentration of C

Molal concentration = moles of solute/kilograms of solvent

Moles of C = 0.187 mol

Mass of A = 47.22 g = 0.047 22 kg

$\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}$

4 0

1 year ago

Dimethylhydrazine is a carbon-hydrogen-nitrogen compound used in rocket fuels. When burned in an excess of oxygen, a 0.312 gg sa

Stolb23 [73]

Answer:

CH₄N

Explanation:

Given that;

mass of the sample = 0.312 g

mass of CO2 = 0.458 g

mass of H2O = 0.374 g

nitrogen content of a 0.486 gg sample is converted to 0.226 gg N2N2.

Let start with calculating the respective numbers of moles of Carbon Hydrogen and Nitrogen from the given data.

numbers of moles of Carbon from CO2 = $\frac{mass of CO_2}{molarmass}*\frac{1 mole of C}{1 mole of CO_2}$

= $\frac{0.458}{44}*\frac{1mole of C}{1 mole of CO_2}$

= 0.0104 mole

numbers of moles of hydrogen from H2O = $\frac{mass of H_2O}{molarmass}*\frac{2 mole of H}{1 mole of H_2O}$

= $\frac{0.374}{18.02}*\frac{2 mole of H}{1 mole of H_2O}$

= 0.02077 × 2

= 0.0415 mole

The nitrogen content of a 0.486 g sample is converted to 0.226 g N2

Now, in 1 g of the sample; The nitrogen content = $\frac{0.226}{0.486}*1$

in 0.312 g of the sample, the nitrogen content will be; $\frac{0.226}{0.486}*0.312$

= 0.1450 g of N2

number of moles of N2 = $\frac{mass}{molar mass}* \frac{2 mole}{1 mole}$

= $\frac{0.1450}{28.0134} *\frac{2 mole}{1 mole}$

= 0.0103 mole

Finally to determine the empirical formula of Carbon Hydrogen and Nitrogen; we have:

Carbon Hydrogen Nitrogen

number of moles 0.0104 0.0415 0.0103

divided by the

smallest number $\frac{0.0104}{0.0103}$ $\frac{0.0415}{0.0103}$ $\frac{0.0103}{0.0103}$

of moles

1 : 4 : 1

∴ The empirical formula = CH₄N

8 0

1 year ago

if a sample of gas is intially at 1.8 atm,22.0 l, and 26.4 c, what will be the volume if the pressure is reduced by 0.8 atm and

Tems11 [23]

pv=nrT Initial state (1.8atm)(22.0 l)=n(0.082057)(26.4+273.15); r=.082057, and converting C to K Solving for n = (1.8)(22)/(.082057*(26.4+273.15) moles n = 1.611 moles in initial state Now we solve for new volume pv=nrT (.8atm)v=(1.611)(.082057)(20.3+273.15) v=(1.611)(.082057)(20.3+273.15)/.8 v=48.49 l

8 0

1 year ago

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