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2 years ago

How many moles of silver chloride, produced from 100 g of silver nitrate reacting with barium chloride BaCl2?

Chemistry

1 answer:

Dmitrij [34]2 years ago

7 0

Answer:

Equation of Reaction

2AgNO3 + BaCl2 === 2AgCl + Ba(NO3)2

Molar Mass of AgNO3 = 170g/mol

Moles of reacting AgNO3 = 100g/170gmol-¹

=0.588moles of AgNO3

From the equation of reaction...2moles of AgNO3 reacts to Produce 2Moles of Silver Chloride

So Their ratio is 2:2.

This means that 0.588Moles of AgCl Will be produced too.

ANSWER...0.588MOLES OF AgCl WILL BE PRODUCED.

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A sample of gas occupies 280 mL when the pressure is 560.00 mm Hg . If the temperature remains constant , what is the new pressu

vichka [17]

Answer : The new pressure if the volume changes to 560.0 mL is, 280 mmHg

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 560.00 mmHg

$P_2$ = final pressure = ?

$V_1$ = initial volume = 280 mL

$V_2$ = final volume = 560.0 mL

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

$560.00mmHg\times 280 mL=P_2\times 560.0 mL$

$P_2=280mmHg$

Therefore, the new pressure if the volume changes to 560.0 mL is, 280 mmHg

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2 years ago

When a lithium ion and nitrate ion combine, the result is:

maxonik [38]

LiNO3; nitrate has a charge of -1, and lithium has a charge of +1

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2 years ago

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molar mass of beryllium (Be) molar mass of silicon (Si) molar mass of calcium (Ca) molar mass of rhodium (Rh)

Neko [114]

Molarmass of beryllium is 9.0
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2 years ago

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A sample of oxygen gas is compressed from 30.6 L to 1.8 L at constant temperature pressure of 1.8 atm. Calculate the amount of e

ad-work [718]

Answer:

the change in the internal energy of the system is 3,752.67 J

Explanation:

Given;

initial volume of the gas, V₁ = 30.6 L

final volume of the gas, V₂ = 1.8 L

constant pressure of the gas, P = 1.8 atm

Energy released by the system, Q = 1.5 kJ = 1,500 J

Apply pressure-volume work equation, to determine the work done on the gas;

w = -PΔV

w = -P(V₂ - V₁)

w = - 1.8 atm(1.8 L - 30.6 L)

w = 51.84 L.atm

w = 51.84 L.atm x 101.325 J/L.atm

w = 5,252.67 J

The change in the internal energy of the system is calculated as;

ΔU = Q + w

Since the heat is given out, Q = - 1,500 J

ΔU = -1,500 J + 5,252.67 J

ΔU = 3,752.67 J

Therefore, the change in the internal energy of the system is 3,752.67 J

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2 years ago