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4 years ago

What is 80% as a ratio

Chemistry

1 answer:

Rashid [163]4 years ago

7 0

Answer:

4 : 5

Explanation:

80 percent can be written as:

80/100

Now we can reduce the above fraction to its lowest term as shown below

80/100 = 8/10

Divide both the numerator and denominator by 2.

8/10 = 4/5

Now, we can not divide further. So the ratio becomes 4 : 5

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1. _ LiBr2 + __K2CO3 -->___ LiCO3 +_ KBr Reaction Type:

Arada [10]

Answer:

Double displacement chemical reaction.

Explanation:

Hello,

In this case, for the given reaction we firstly balance it:

$LiBr_2 +K_2CO3 \rightarrow LiCO3 +2KBr$

Then, since the lithium and potassium cations are being exchanged to each other from bromide and carbonate, we are talking about a double displacement chemical reaction.

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5 0

4 years ago

Which is a reactant in the cellular respiration reaction

Alexxandr [17]

Answer: Oxygen and glucose are both reactants in the process of cellular respiration.

Explanation: Oxygen and glucose are both reactants in the process of cellular respiration. The main product of cellular respiration is ATP; waste products include carbon dioxide and water.

7 0

3 years ago

Read 2 more answers

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:

Step 1: 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

Step 2: 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

Step 3: Calculate moles of sodium formate:

Moles sodium formate = molarity * volume

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

Step 4: 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

Write about the chemicals which can bring the permanent hardness in water

iogann1982 [59]

Answer:

Permanent hardness (mineral content) is generally difficult to remove by boiling. If this occurs, it is usually caused by the presence of calcium sulfate/calcium chloride and/or magnesium sulfate/magnesium chloride in the water, which do not precipitate out as the temperature increases

Explanation:

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7 0

2 years ago

I really kinda need some hell right now :)

Fudgin [204]

It’s c 1.59 x 10^4L i’m positive

4 0

3 years ago