CHEMISTRY COLLEGE

How many grams of chloride are there in 256 g of magnesium chloride?

Answers

Answer 1

Answer:

Given :

Mass of given magnesium chloride, m = 256 g.

To Find :

How many grams of chloride are there in 256 g of magnesium chloride.

Solution :

Molecular formula of magnesium chloride is $MgCl_2$ .

Molecular formula of $MgCl_2$ is, M = 95.211 g/mol .

Mass of chlorine in 1 mol of $MgCl_2$ is , m = 35.5 × 2 = 71 g.

So, amount of chlorine in 256 gram $MgCl_2$ :

$m =(71* 256)/(95.211)\n\nm = 190.90\ g$

Hence, this is the required solution.

Related Questions

How many moles of oxygen are in 8.24 moles Mg(NO3)2
A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration of a solution equals the mass of what's dissolved divided by the total volume of the solution. Here's how the student prepared the solution: The label on the graduated cylinder says: empty weight: 4.5 g She put some solid sodium hydroxide into the graduated cylinder and weighed it. With the sodium hydroxide added, the cylinder weighed 48.772 g. She added water to the graduated cylinder and dissolved the sodium hydroxide completely. Then she read the total volume of the solution from the markings on the graduated cylinder. The total volume of the solution was 152.3 mL. What concentration should the student write down in her lab notebook? Be sure your answer has the correct number of significant digits.
Calculate the mass of silver chloride required to plate 265 mg of pure silver.
1. Write the structural formulas and give IUPAC names for all isorneric alcohols of molecular formulaС4Н9OH
What are some examples of matter?

A natural atom possesses the atomic number of 13 and a atomic mass of 27. Three electrons are lost. From what region of the atom are they lost?

Answers

Answer:

The electrons are lost from the valence shell (outermost electron shell) of the atom.

Explanation:

This is able to be inferred not only because valence electrons being lost first is a trend but also because the atom in question has actually 3 valence electrons (13-2-8 = 3).

What element is found in all organic compounds?
a.carbon?

Answers

The element that is found in all organic compounds is Carbon

9. The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0120 mol gaseous NH3 and 0.0170 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 2.20×10-3 M. Calculate Keq for the reaction at this temperature.

Answers

Answer: The value of $K_(eq)$ is $4.66* 10^(-5)$

Explanation:

We are given:

Initial moles of ammonia = 0.0120 moles

Initial moles of oxygen gas = 0.0170 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $(0.0120)/(1.00)=0.0120M$

Concentration of oxygen gas = $(0.0170)/(1.00)=0.0170M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0120 0.0170

At eqllm: 0.0120-4x 0.0170-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $2.20* 10^(-3)M=0.00220M$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.00220\n\n\Rightarrow x=(0.00220)/(2)=0.00110M$

Now, equilibrium concentration of ammonia = $0.0120-4x=[0.0120-(4* 0.00110)]=0.00760M$

Equilibrium concentration of oxygen gas = $0.0170-3x=[0.0170-(3* 0.00110)]=0.0137M$

Equilibrium concentration of water = $6x=[6* 0.00110]=0.00660M$

The expression of $K_(eq)$ for the above reaction follows:

$K_(eq)=([N_2]^2* [H_2O]^6)/([NH_3]^4* [O_2]^3)$

Putting values in above expression, we get:

$K_(eq)=((0.00220)^2* (0.00660)^6)/((0.00760)^4* (0.0137)^3)\n\nK_(eq)=4.66* 10^(-5)$

Hence, the value of $K_(eq)$ is $4.66* 10^(-5)$

Which of the following shows the abbreviation for an SI unit of density?

Answers

Is there supposed to be a graph or something?

the aswer is g/ml hope is helpful

Given that e = 9.0 v , r = 98 ω and c = 23 μf , how much charge is on the capacitor at time t = 4.0 ms

Answers

Let charge across the capacitor be Q, current through the circuit be I.
Voltage difference across the resistor = rI
Voltage difference across the capacitor = Q/c
Loop rule: net voltage change through a loop must be zero, so
9 = rI + Q/c. Since I = dQ/dt,
r dQ/dt + Q/c = 9
Solving, Q = 9c (1 - e^(t/rc)). Plug in the numbers from the problem for the numerical answer.

Determine the rate of a reaction that follows the rate law:rate = k[A]”[B]", where:
k= 1.5
[A] = 1 M
[B] = 3 M
m = 2
n = 1

Answers

The rate of the reaction is 4.5 mol L⁻¹s⁻¹.

What is meant by rate of a reaction ?

Rate of a reaction is defined as the change in concentration of any one of the reactants or products of the reaction, in unit time.

Here,

The concentration of A, [A] = 1 M

The concentration of B, [B] = 3 M

The partial order with respect to A, m = 2

The partial order with respect to B, n = 1

The rate constant of the reaction, k = 1.5

The rate of the reaction,

r = k[A]^m [B}^n

r = 1.5 x 1² x 3

r = 4.5 mol L⁻¹s⁻¹

Hence,

The rate of the reaction is 4.5 mol L⁻¹s⁻¹.

To learn more about rate of a reaction, click:

brainly.com/question/29049197

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Answer:

k= 1.5

[A] = 1 M

[B] = 3 M

m = 2

n = 1

Explanation:

rate = k[A]”[B]"

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