Ask question

Ask question

All categories

1 year ago

13

Calculate the molarity of a 10. 0% cacl₂ solution. The density of the solution is 1. 0835 g/cm³.

1 answer:

brilliants [131]1 year ago

5 0

The molarity of 10% CaCl2 is 0.9%

concentration of the given salt CaCl₂ = 10%

Density of a solution = 1.0835 g/cm³

Volume = m / d

= 100 / 1.0835

= 92.29 litres

Density = mass / volume

1.0835 × 92.29 = mass

mass = 99.99 gram

Thus the molarity can be calculated by = moles of solute / volume of solution multiplied by 100

= 0.9008/ 92.29 X 100 %

= 0.009 X 100 %

= 0.9 %

The molarity of 10% CaCl2 is 0.9%

To know more about density and molarity you may visit the link which is mentioned below:

brainly.com/question/10710093

#SPJ4

You might be interested in

The charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electro

shusha [124]

Answer:

$1.6\times 10^{-18} C$

Explanation:

The fundamental charge of a single electron is $1.6\times 10^{-19} C$ .

If there are 10 excess electrons, the net charge that would be measured should be 10 times the fundamental charge of a single electron:

$Q=nq_e\\Q= 10\times 1.6\times 10^{-19} C\\Q= 1.6 \times 10^{-18} C$

3 0

3 years ago

At what water temperature will additional heat energy need to be added before the temperature will change again?

Lostsunrise [7]

That would be 0 degrees Celsius aka the melting point of water.... If you look at the diagram I attached you notice that at 0 degrees Celsius it is flat, this is because much heat is needed at this point for water to rise to 1 degree... It is the same for the boiling point (100)<span />

6 0

3 years ago

Is noise energy a kind of energy or not

Volgvan

Yes. sound waves are produced by energy.

4 0

3 years ago

Experimenting with free fall, Mariana observes that her baseball takes 1.5 s to travel the last 30m before hitting the ground. F

Art [367]

Answer:

37.8 m

Explanation:

At point 0, the ball is at height y₀.

At point 1, the ball is at height 30 m.

At point 2, the ball is at height 0 m.

Given:

y₁ = 30 m

y₂ = 0 m

v₀ = 0 m/s

a = -10 m/s²

t₂ − t₁ = 1.5 s

Find: y₀

Use constant acceleration equation.

y = y₀ + v₀ t + ½ at²

Evaluate at point 1.

y₁ = y₀ + v₀ t₁ + ½ at₁²

30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²

30 = y₀ − 5t₁²

Evaluate at point 2.

y₂ = y₀ + v₀ t₂ + ½ at₂²

0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²

0 = y₀ − 5t₂²

y₀ = 5t₂²

Substitute:

y₀ = 5 (1.5 + t₁)²

y₀ = 5 (2.25 + 3t₁ + t₁²)

y₀ = 11.25 + 15t₁ + 5t₁²

30 = 11.25 + 15t₁ + 5t₁² − 5t₁²

30 = 11.25 + 15t₁

t₁ = 1.25

30 = y₀ − 5t₁²

30 = y₀ − 5(1.25)²

y₀ ≈ 37.8

4 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 27 ft/s. Its height

Ahat [919]

Answer:

$V_{3.01}=-93.2m/s$

$V_{3.005}=-93.1m/s$

$V_{3.002}=-93.04m/s$

$V_{3.001}=-93.02m/s$

$V_{3}=-93m/s$

Explanation:

To calculate average velocity we need the position for both instants t0 and t1.

Now we will proceed to calculate all the positions we need:

$Y_{3}=-99m/s$

$Y_{3.01}=-99.932m/s$

$Y_{3.005}=-99.4655m/s$

$Y_{3.002}=-99.18608m/s$

$Y_{3.001}=-99.09302m/s$

Replacing these values into the formula for average velocity:

$V_{3-3.01}=\frac{Y_{3.01}-Y_{3}}{3.01-3}=-93.2m/s$

$V_{3-3.005}=\frac{Y_{3.005}-Y_{3}}{3.005-3}=-93.1m/s$

$V_{3-3.002}=\frac{Y_{3.002}-Y_{3}}{3.005-3}=-93.04m/s$

$V_{3-3.001}=\frac{Y_{3.001}-Y_{3}}{3.001-3}=-93.02m/s$

To know the actual velocity, we derive the position and we get:

$V=27-40t = -93m/s$

5 0

3 years ago