All categories

3 years ago

I will mark brainlist please help

Chemistry

1 answer:

Lostsunrise [7]3 years ago

7 0

Answer: 20) 2365 g

21) 22.39 grams.

22) 29.99 g

Explanation: 20) molarity is the no of moles of solute per unit volume.

We can calculate amount of CaCl2 required to prepare 0.1 M CaCl2 1000 ml solution.

we know that to prepare one ltr of 1 M solution of CaCl2 111 g required

Now consider x gram will require to prepare to

so that comparing above both condition

1000ml ×1M×X g=1000ml×0.1M×111g

X= 11.1 gram

X= 11.1 g of CaCl2

Hence 11.1 g of CaCl2 would be dissolved in 1.0L of a 0.100 M solution of CaCl2

21) How many moles of CaCl₂ in that solution?

;

What's the mass of that 0.20172 moles of CaCl₂?

Molar mass from a modern periodic table:

Ca- 40.078;

Cl- 35.45.

Molar mass of CaCl₂:

Mass of that 0.20172 moles of CaCl₂:

22) its a 3.0m solution so 1 litre of solution contains 3 moles of NaOH, 250ml of solution contains 0.25x39.9971 g/mol, so 250ml of this solution contains 0.75x39.9971=29.99g, or if you round it up 30.0g

You might be interested in

How many moles are in 54.0 grams of Silver?

Lelechka [254]

54g ag *(108mol ag/1 g ag) =5832mol ag

4 0

4 years ago

Please help asap thank you.

Aleks [24]

Answer:

a)0.816g/l

b)0.241g/l

8 0

3 years ago

If the reaction N2 (g) + 3 H2 (g) --> 2 NH3 (g) has the concentrations 1.1 M for nitrogen, 0.75 M for hydrogen and 0.25 M fo

Luba_88 [7]

<u>Answer:</u> The value of $K_c$ is 0.136 and is reactant favored.

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For the chemical reaction between carbon monoxide and hydrogen follows the equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

The expression for the $K_{c}$ is given as:

$K_{c}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

We are given:

$[NH_3]=0.25M$

$[H_2]=0.75M$

$[N_2]=1.1M$

Putting values in above equation, we get:

$K_c=\frac{(0.25)^2}{1.1\times (0.75)^3}$

$K_c=0.135$

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.
When $K_{c}$ ; the reaction is reactant favored.
When $K_{c}=1$ ; the reaction is in equilibrium.

For the given reaction, the value of $K_c$ is less than 1. Thus, the reaction is reactant favored.

Hence, the value of $K_c$ is 0.136 and is reactant favored.

4 0

3 years ago

In an atom, the number of neutrons determines most specifically the ________.

lord [1]

In an atom, the number of neutrons determines most specifically the isotopes of an element.

<h2>Further Explanation</h2><h3>An atom </h3>

An atom is defined as the smallest particle of an element that can take part in a chemical reaction.
Atoms are made of energy shells and the inner nucleus.
It is also made up of subatomic particles, protons and neutrons.

<h3>Neutrons </h3>

They are sub-atomic particles in an atom that are found in the nucleus of an atom. They have a zero charge or no charge.
Neutrons are slightly heavier than protons.
The number of protons and neutrons in a nucleus of an atom results to the atomic mass number of an atom. Like protons they only take part in nuclear reactions.

<h3>Protons</h3>

Protons are subatomic particles that are found in the nucleus of an atom. They are positively charged. with a charge of +1.
The number of protons differ from one element to another and thus used as the atomic number of an individual element.
The nucleus of an atom is positively charged due to the positive charge of the protons.
Additionally, protons do not take part in chemical reactions but in nuclear reactions.

<h3>Electrons </h3>

They are found orbiting the nucleus on energy shells based on Clouds’ theory. Electrons are negatively charged each with a charge of -1.
The number of electrons and protons in a neutral atom are equal. Additionally they have the least mass of the three sub-atomic particles.
Electrons are the only sub-atomic particles that take part in chemical reactions.

<h3>Isotopes </h3>