1. K + Cl2 —> KCI
Given: 180 g K and 120 g CI2

1. K + Cl2 > KCI Given: 180 G K And 120 G CI2

Answers

Answer 1

Answer:

A.  Cl₂

B.  51.36 g K

C.  245.3 g KCl

Explanation:

Balance the given chemical equation

K  +  Cl₂  ⇒  KCl

2 K  +  Cl₂  ⇒  2 KCl

A. To find the limiting reagent, first convert the mass of each reactant to moles.  The molar masses are 39.10 g/mol for K and 70.91 g/mol for Cl₂.

(180 g)/(39.10 g/mol) = 4.60 mol K

(120 g)/(70.91 g/mol) = 1.69 mol Cl₂

Now, convert moles of each reactant to moles of the product.  You can do this by using the mole ratio in the balanced chemical reaction.  The reactant with the smallest moles of product is the limiting reagent.

(4.60 mol K) × (2 mol KCl/2 mol K) = 4.60 mol KCl

(1.69 mol Cl₂) × (2 mol KCl/1 mol Cl₂) = 3.29 mol KCl

Cl₂ is the limiting reagent.

B.  You need to find out how much excess K you will have.  Since Cl₂ is the limiting reagent, you will only make 3.29 mol of KCl.  Convert this value to moles of K using the mole ratio.

(3.29 mol KCl) × (2 mol K/2 mol KCl) = 3.29 mol K

Now convert moles of K to grams.  The molar mass is 39.10 g/mol.

(3.29 mol K) × (39.10 g/mol) = 128.64 g K

This is how much K will be consumed in the reaction.  Subtract the total amount of K by the amount that will be consumed.

180 - 128.64 = 51.36 g

There will be 51.36 g of excess K.

C.  You know that you will produce 3.29 mol of KCl based on previous calculations.  Convert this to grams.  The molar mass is 74.55 g/mol.

(3.29 mol) × (74.55 g/mol) = 245.3 g KCl

You will produce 245.3 g of KCl.


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Answers

Answer: C. Squamous

Explanation:

I got it right on my test...lol

The type of epithelial cells that would be found in an area of the body where diffusion of nutrients needs to occur is squamous.

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Answers

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2Ni(H₃O₄)₂

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The given formula is actually having two moles of Ni(H₃O₄)₂.

Thus in two mole total number of atoms are = 2+12+16 =30 atoms

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Answers

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Answers

Answer:

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Explanation:

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Question 14 On Mars, the element zircon has two isotopes, Zr156 and Zr160.84% of a sample is Zr156, which has an atomic mass of 156.8 amu, and 16% Zr160, which has an atomic mass of 159.4 amu. What is the average atomic mass of zircon on Mars? 157.2 amu 158.0 amu 159.2 amu 162.4 amu​

Answers

Answer:

Option A is correct = 157.2 amu

Explanation:

Given data:

Percentage abundance of Zr-156 = 84%

Atomic mass of Zr-156 = 156.8 amu

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Answers

Answer:

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Answer:

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Refer to above attachment...

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Answers

Answer:

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Explanation:

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From the question

actual density = 8.96 g/mL

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We have

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Answers

Explanation:

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Answers

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Explanation:

The density of lead is 11.4 g/cm³

We need to convert the density in kg/m³ in scientific notation.

We know that,

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Answers

Answer:

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Explanation:

aluminum is a metal.

potassium is a metal.

calcium is a metal.

But chlorine is a nonmetal.

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