All categories

3 years ago

When 80.0 grams of calcium chloride react with silver nitrate, how many grams of silver

Chemistry

1 answer:

hichkok12 [17]3 years ago

5 0

Answer:

Mass = 206.38 g

Explanation:

Given data:

Mass of calcium chloride = 80.0 g

Mass of silver chloride formed = ?

Solution:

Chemical equation:

CaCl₂ + 2AgNO₃ → Ca(NO₃)₂ + 2AgCl

Number of moles of CaCl₂:

Number of moles = mass/molar mass

Number of moles = 80.0 g/ 110.98 g/mol

Number of moles = 0.72 mol

Now we will compare the moles of CaCl₂ and AgCl

CaCl₂ : AgCl

1 : 2

0.72 : 2/1×0.72 = 1.44 mol

Mass of silver chloride:

Mass = number of moles × molar mass

Mass = 1.44 mol × 143.32 g/mol

Mass = 206.38 g

You might be interested in

. Ashley is identifying insects and finds a camo moth. A camo moth appears like a leaf and is able to easily blend in with its s

In-s [12.5K]

I think the answer is C

5 0

3 years ago

Read 2 more answers

Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have

Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

7 0

3 years ago

What is the area of triangle ABC with vertices A(x¹,y¹), B(x²,y²)and C (x³,y³)??????????<br>

AlexFokin [52]

Answer:

$Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|$

$Area = 2\ units^2$

Explanation:

Given

$A = (x_1,y_1)$

$B = (x_2,y_2)$

$C = (x_3,y_3)$

Required

Determine the area

The area of a triangle is :

$Area = \frac{1}{2}|A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)|$

By substituting values for the x and y coordinates of A, B and C;

We have:

$Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|$

So:

For instance

$A = (0,3)$

$B= (2,1)$

$C = (2,3)$

The area is:

$Area = \frac{1}{2}|0(1-3) + 2(3-3) + 2(3-1)|$

$Area = \frac{1}{2}| 2*0 + 2*2|$

$Area = \frac{1}{2}| 0 + 4|$

$Area = \frac{1}{2}|4|$

$Area = \frac{1}{2} * 4$

$Area = 2\ units^2$

7 0

3 years ago

The true absorbance for a 1.0 x 10 −5 M solution is 0.7526. If the percentage stray light for a spectrophotometer is 0.56%, calc

Korvikt [17]

Answer:

The percentage deviation is $\Delta M = 1.87$ %

Explanation:

From the question we are told that

The concentration is of the solution is $C = 1.0*10^{-5} M$

The true absorbance A = 0.7526

The percentage of transmittance due to stray light $z = 0.56$ % $=\frac{0.56}{100} = 0.0056$

Generally Absorbance is mathematically represented as

$A = -log T$

Where T is the percentage of true transmittance

Substituting value

$0.7526 = - log T$

$T = 10^{-0.7526}$

$= 0.177$

$= 17.7$ %

The Apparent absorbance is mathematically represented

$A_p = -log (T +z)$

Substituting values

$A_p = -log(0.177 + 0.0056)$

$= -log(0.1826)$

= 0.7385

The percentage by which apparent absorbance deviates from known absorbance is mathematically evaluated as

$\Delta A = \frac{A -A_p}{A} * \frac{100}{1}$

$= \frac{0.7526 - 0.7385}{0.7526} * \frac{100}{1}$

$\Delta A = 1.87$ %

Since Absorbance varies directly with concentration the percentage deviation of the apparent concentration from know concentration is

$\Delta M = 1.87$ %

6 0

4 years ago

A gas has experienced a small increase in volume but has maintained the same pressure and number of moles. According to the idea

krek1111 [17]

Answer:

If a gas has experienced a small increase in volume but has maintained the same pressure and number of moles, the temperature of the gas will DROP.

Explanation:

According to Boyle’s law of ideal gases, volume and temperature of a gas is inversely related, as long as the pressure is kept constant;

P₁V₁/T₁ = P₂V₂/T₂

Therefore, if the volume of the gas increases, the temperature will definitely decrease due to the inverse relationship. The gas will get cooler.

Learn More:

For more on Boyle's Law check out;

brainly.com/question/13362447

brainly.com/question/2568628

brainly.com/question/12049334

#LearnWithBrainly

4 0

3 years ago