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Ann [662]
3 years ago
8

You have a square with a Perimeter of 16 in,, what is the Area of the square after scaling it by 5?​

Chemistry
1 answer:
Novosadov [1.4K]3 years ago
8 0

Answer:

f the figure is a square with a perimeter of 16 inches, then each side of the square is 4 inchest in length. Area is found by multiplying the length x the width. For a square, the length and width are the same. A = 16 sq

Explanation:

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You have 150. g of a bleach solution. The percent by mass of the solute, sodium hypochlorite (NaOCI), is 3.62%.
Taya2010 [7]

Answer:

There are 5.43 grams of NaOCl

Explanation:

The given percent by mass of the solute tells us that out of the 150 g of the solution, 3.62% are due solely to the solute.

In other words, <u>the mass of the solute in the solution is</u>:

  • 150 g * 3.62/100 = 5.43 g

Thus, in 150 grams of the given bleach solution, there are 5.43 grams of sodium hypochlorite.

4 0
2 years ago
I need a data and observation for balls and how many times they bounce or whatever
Allisa [31]
Conduct experiments to see which hall Bounce the higher and see how they could collect data
8 0
2 years ago
What mass of solid sodium formate (of MW 68.01) must be added to 150 mL of 0.42 mol/L formic acid (HCOOH) to make a buffer solu-
Sergio [31]

Answer:

We need 4.28 grams of sodium formate

Explanation:

<u>Step 1:</u> Data given

MW of sodium formate = 68.01 g/mol

Volume of 0.42 mol/L formic acid = 150 mL = 0.150 L

pH = 3.74

Ka = 0.00018

<u>Step 2:</u> Calculate [base)

3.74 = -log(0.00018) + log [base]/[acid]

0 = log [base]/[acid]

0 = log [base] / 0.42

10^0 = 1 = [base]/0.42 M

[base] = 0.42 M

<u>Step 3:</u> Calculate moles of sodium formate:

Moles sodium formate = molarity * volume

Moles of sodium formate = 0.42 M * 0.150 L = 0.063 moles

<u>Step 4:</u> Calculate mass of sodium formate:

Mass sodium formate = moles sodium formate * Molar mass sodium formate

Mass sodium formate = 0.063 mol * 68.01 g/mol

Mass sodium formate = 4.28 grams

We need 4.28 grams of sodium formate

4 0
3 years ago
Calculate ΔG o for the following reaction at 25°C: 3Mg(s) + 2Al3+(aq) ⇌ 3Mg2+(aq) + 2Al(s) Enter your answer in scientific notat
larisa [96]

Answer:

-3.7771 × 10² kJ/mol

Explanation:

Let's consider the following equation.

3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

n: moles

ΔG°f(): standard Gibbs free energy of formation

p: products

r: reactants

ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))

ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)

ΔG° = -377.71 kJ = -3.7771 × 10² kJ

This is the standard Gibbs free energy per mole of reaction.

5 0
2 years ago
Why can gasses change volume?​
soldi70 [24.7K]

Answer:

b. I've seen a question like this before lol

7 0
3 years ago
Read 2 more answers
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