Who's 10th grade in CA Chemistry?

Which phase change is an example of an exothermic process?

Ivenika [448]

It could be either sublimation, freezing or condensation. If i had more information i could tell you which

3 0

4 years ago

Which of the following would be a clue that a rock is igneous?

ryzh [129]

Answer:

A

Explanation:

they usually have that Crystal like thing in them... since they are formed from Lava

7 0

3 years ago

Read 2 more answers

How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __

kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

determine the number of moles of HCl
determine the mole ratio,
use the mole ratio to calculate the number of moles of aluminum.
use RFM of Aluminum to determine the grams required.

Moles of HCl

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

$x \: mol \: = \: \frac{2 \: \times \: 35}{1000} \\ = 0.07 \: moles \:$

We have 0.07 moles of HCl.

Mole ratio

Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

but moles of HCl is 0.07, therefore the moles of Al;

$= \frac{2}{6} \times 0.07 \\ \: = 0.0233333 \: moles$

Therefore we have 0.0233333 moles of aluminum.

Grams of Aluminum

We use the formula;

$grams \: = moles \: \times \: rfm$

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

$= 0.0233333 \: moles \: \times \: 26.982 \: \frac{g}{mol} \\ = 0.625799 \: grams$

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0

2 years ago

2.A calibration curve requires the preparation of a set of known concentrations of CV, which are usually prepared by dieting a s

Aleks04 [339]

Answer:

In order to prepare 10 mL, 5 μM; 2 mL of the 25 μM stock solution will be taken and diluted with water up to 10 mL mark.

In order to prepare 10 mL, 10 μM; 4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 15 μM; 6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 20 μM; 8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

Explanation:

Using the dilution equation:

no of moles before dilution = no of moles after dilution.

Molarity x volume (initial)= Molarity x volume (final).

In order to prepare 10 mL, 5 μM from 25 μM solution,

Final molarity = 5 μM, final volume = 10 mL, initial molarity = 25 μM, initial volume = ?

25 x initial volume = 5 x 10

Initial volume = 50/25

= 2 mL

2 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 10 μM from 25 μM stock,

25 x initial volume = 10 x 10

Initial volume = 100/25 = 4 mL

4 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 15 μM from 25 μM stock,

25 x initial volume = 15 x 10

initial volume = 150/25 = 6 mL

6 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

In order to prepare 10 mL, 20 μM from 25 μM stock,

25 x initial volume = 20 x 10

initial volume = 200/25 = 8 mL

8 mL of the 25 μM stock solution will be taken and diluted up to 10 mL mark.

6 0

3 years ago

if three oxygen particles are needed to form ozone, how many units of ozone could be formed from 6 oxygen particles? from 9? fro

tensa zangetsu [6.8K]

<h3>Answers:</h3>

1) 2 Units of Ozone

2) 3 Units of Ozone

3) 9 Units of Ozone

<h3>Solution:</h3>

1) From 6 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

6 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 2 Units of Ozone

2) From 9 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

9 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 3 Units of Ozone

3) From 27 Oxygen Particles;

As given,

3 Oxygen Particles form = 1 Unit of Ozone

So,

27 Oxygen Particles will form = X Units of Ozone

Solving for X,

X = (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

X = 9 Units of Ozone

3 0

3 years ago