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3 years ago

BaCl2 (aq) + AgNO3 (aq) →

Chemistry

2 answers:

sweet [91]3 years ago

8 0

Answer:

BaCl2 + 2AgNO3 —> 2AgCl + Ba(NO3)2

Explanation:

dlinn [17]3 years ago

3 0

Answer:

A white precipitate is formed

Explanation:

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Common uses for halogen compounds

olga2289 [7]

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2 years ago

How do very dry areas of the world become habitable

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3 years ago

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Ksju [112]

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3 years ago

If 20 mL of gas is subjected to a temperature change from 10°C to 100°C and a pressure change from 1 atm to 10 atm,

Evgesh-ka [11]

Answer:

E. None of these

Explanation:

We know, By GAS laws,

PV = NRT, where p- pressure, v- volume, n- number of moles, R- gas constant ,and T- temperature

Now, In the question, the number of moles remains the same as the gas is the same. so n is constant so we can compare n before and after a temperature change.

$\frac{P1V1}{RT1}$ = $\frac{P2V2}{RT2}$

where P1= 1 atm, P2 = 10 atm, V1= 20 mL, T1= 10°C and T2= 100°C

We don't have to worry about the standard units as they are present equally on both the sides and get cut, same goes for R( gas constant)

So putting values, we get

$\frac{1*20}{R*10} = \frac{10*V2}{R*100}$

Cutting, R on both sides and moving contents to the right so that only V2 is left on the left.

$\frac{1*20*100}{10*10} = V2$

∴ V2 = $\frac{2000}{100}$

∴ V2 = 20mL

3 0

2 years ago

The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon a

DIA [1.3K]

Answer:

Empirical formula = C5H4

Molecular formula = C10H8

Explanation:

When the 3000 mg of naphthalene are burned they produce 10.3 mg of CO2. Knowing the unbalanced equation of the combustion of naphthalene, we have:

CxHy + O2 = CO2 + H2O

We calculate the molar composition of the sample. We look for the molecular weights in the periodic table:

CO2 = 12,011 + 2 (15,999) = 44,009 g

Mol C = 10.3 mg * (1 mol CO2 / 44.009 g CO2) * (1 mol C / 1 mol CO2) = 0.234 mmol C

Mass C = 0.234 mmol C * (12.011 g C / 1 mol C) = 2.8105 mg C

Mass H = 3 mg - 2.8105 mg = 0.1895 mg H

Mol H = 0.1895 mg H * (1 mol H / 1,008 g H) = 0.188 mmol H

To calculate the empirical formula, we must divide the number of moles of each element by the smallest number of moles, in this case, of hydrogen:

C = 0.2340 mmol C / 0.1895 mol H = 1.25

H = 0.1895 mmol H / 0.1895 mmol H = 1

We multiply the coefficients by 4, and we have the empirical formula:

C1.25 * 4H1 * 4 = C5H4

The molecular formula is equal to (C5H4)m, where m is calculated by the molecular and empirical mass ratio, as follows:

Empirical mass = (5 * 12.011) + (4 * 1.008) = 64.09 g

m = 130 g / 64.09 g = 2.02 = 2

Therefore we have the molecular formula:

(C5H4)2 = C10H8

4 0

3 years ago