Concentration of OH- for Ca(OH)

A rock that weighs 12.2 g is added to a graduated cylinder with 30.0 mL of water in it. The meniscus rises to 54.4 mL. What is t

GuDViN [60]

Answer:

0.50 g/mL

Explanation:

Formula of Density,

D = mass / volume

Mass = 12.2 g

Volume = 54.4 mL - 30.0 mL = 24.4 mL

Putting values,

D = 12.2 g / 24.4 mL

D = 0.50 g/mL

4 0

2 years ago

The process to separate the various components of a liquid mixture is called a fractional distillation true false

Black_prince [1.1K]

DecantationDecantation is a process for the separation of mixtures, by removing a layer of liquid, generally one from which a precipitate has settled. The purpose may be either to produce a clean decant, or to remove undesired liquid from the precipitate (or other layers). Separate liquids
so it is false

4 0

2 years ago

Read 2 more answers

If the mass of a certain block is 500g, what is the weight of the block?

Dafna11 [192]

______________________________

Solution,

Mass=500g =500/1000=0.5 kg

gravity(g)=9.8m/s^2

Now,

Weight= m*g

 = 0.5*9.8

 =4.9 N

So the weight of the block is 4.9 Newton

hope it helps...

Good luck on your assignment

_______________________

5 0

3 years ago

Why doesn't the earth,s shadow get cast onto the moon every month ?

soldi70 [24.7K]

Becuz the sun refelt off the earth and on to the moon.

8 0

2 years ago

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$

4 0

2 years ago