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Agata [3.3K]
3 years ago
14

You are given a sample of a Group II metal chloride, but the identity of the metal is unknown. Knowing that all Group II metal c

hloride compounds have the general formula, MCI2 and that the percentage of chlorine in the compound is 34.1%, what is the identity of the unknown substance?​
Chemistry
1 answer:
topjm [15]3 years ago
4 0

Answer:

The identity of unknown substance is Barium.

Explanation:

The given compound has a formula of MCl2. And the percentage of chlorine in this compound is said to be 34.1 %. But, we know that the mass of chlorine in this compound is, 35.5*2 = 71 g

Therefore, if we let x be the total mass of the compound, then:

34.1% of x = 71 g

(0.341)x = 71 g

x = 71 g/0.341

x = 208.2 g

Hence, the mass of other atom M, must be:

M = x - 71 g

M = 208.2 g - 71 g

M = 137.2 g

Now, we look into periodic table in group II. We find that the element is Barium with atomic mass of 137 g.

<u>The identity of unknown substance is Barium.</u>

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As we know that

P.E. = mgh

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m= mass of object= 3kg

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h = height between object and animal = 0 m

Then

P.E. = 3× 9.8 × 0 = 0 Joules or 0J

<em>Have a luvely day!</em>

6 0
2 years ago
Anyone could help me out?
Hoochie [10]

1.4 mg/dL = 0.014 g/L

Explanation:

Milligrams per deciliter to grams per liter

There is 1000 grams of mg/dL of 1 g/L

8 0
2 years ago
Read 2 more answers
Calculate the radius ratio for NaBr if the ionic radii of Na + and Br − are 102 pm and 196 pm , respectively. radius ratio: Base
Fudgin [204]

Answer : The expected coordination number of NaBr is, 6.

Explanation :

Cation-anion radius ratio : It is defined as the ratio of the ionic radius of the cation to the ionic radius of the anion in a cation-anion compound.

This is represented by,

\frac{r_{cation}}{r_{anion}}

When the radius ratio is greater than 0.155, then the compound will be stable.

Now we have to determine the radius ration for NaBr.

Given:

Radius of cation, Na^+ = 102 pm

Radius of cation, Br^- = 196 pm

\frac{r_{cation}}{r_{anion}}=\frac{102}{196}=0.520

As per question, the radius of cation-anion ratio is between 0.414-0.732. So, the coordination number of NaBr will be, 6.

The relation between radius ratio and coordination number are shown below.

Therefore, the expected coordination number of NaBr is, 6.

8 0
3 years ago
What is the molarity of a NaOH solution when 22.14 mL of 0.105 M oxalic acid is needed to
jeka57 [31]

The molarity of the NaOH solution required for the reaction is 0.186 M

We'll begin by writing the balanced equation for the reaction

H₂C₂O₄ +2NaOH —> Na₂C₂O₄ + 2H₂O

From the balanced equation above,

The mole ratio of the acid, H₂C₂O₄ (nA) = 1

The mole ratio of the base, NaOH (nB) = 2

From the question given above, the following data were obtained:

Volume of the acid, H₂C₂O₄ (Va) = 22.14 mL

Molarity of the acid, H₂C₂O₄ (Ma) = 0.105 M

Volume of the base, NaOH (Vb) = 25 mL

<h3>Molarity of the base, NaOH (Mb) =? </h3>

MaVa / MbVb = nA/nB

(0.105 × 22.14) / (Mb × 25) = 1/2

2.3247 / (Mb × 25) = 1/2

Cross multiply

Mb × 25 = 2.3247 × 2

Mb × 25 = 4.6494

Divide both side by 25

Mb = 4.6494 / 25

<h3>Mb = 0.186 M</h3>

Therefore, the molarity of the base, NaOH solution is 0.186 M

Learn more: brainly.com/question/25739717

4 0
2 years ago
16g of O2
cricket20 [7]

they are listed in order of decreasing number of moles

6 0
3 years ago
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