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geniusboy [140]
2 years ago
11

Why are fingertips wet with saliva before counting money?​

Chemistry
1 answer:
poizon [28]2 years ago
8 0

Answer:

It helps to get a better grip on the money.

Explanation:

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A _____ ionic compound is a polyatomic ionic compound composed of three or more different elements.
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Answer:

ternary

Explanation:

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Which gas is the most abundant greenhouse gas?
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The most abundant greenhouse gas is water vapor.

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How do autotrophs (plants) recycle matter?
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Through absorbing nutrients and matter and the releasing new matter, usually in the form of CO2<span />
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3 years ago
Shaun's doctor has recommended that he consume 200 calories of protein a day. Shaun consumed 176 calories of protein in one day.
evablogger [386]

Answer:  Shaun must consume 6 more g of protein to reach the required amount.

Explanation:

Given : Recommended amount of protein = 200 calories

Now 4 calories = 1 gram of protein

Thus 200 calories = \frac{1}{4}\times 200=50 g of protein.

Calories consumed per day by Shaun = 176 calories

176 calories =  \frac{1}{4}\times 176=44 g of protein.

Thus Shaun need to consume = (50-44) g = 6 of protein to reach the required amount.

Shaun must consume 6 more g of protein to reach the required amount.

5 0
2 years ago
An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; M = 62.07 g/mol) and water
maks197457 [2]

Answer :

(a) The volume percent is, 50.63 %

(b) The mass percent is, 52.69 %

(c) Molarity is, 9.087 mole/L

(d) Molality is, 17.947 mole/L

(e) Moles fraction of ethylene glycol is, 0.244

Explanation : Given,

Density of ethylene glycol = 1.114 g/mL

Molar mass of ethylene glycol = 62.07 g/mole

Density of water = 1.00 g/mL

Density of solution or mixture = 1.070 g/mL

According to the question, the mixture is made by mixing equal volumes of ethylene glycol and water.

Suppose the volume of each component in the mixture is, 1 mL

First we have to calculate the mass of ethylene glycol.

\text{Mass of ethylene glycol}=\text{Density of ethylene glycol}\times \text{Volume of ethylene glycol}=1.114g/mL\times 1mL=1.114g

Now we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.00g/mL\times 1mL=1.00g

Now we have to calculate the mass of solution.

Mass of solution = Mass of ethylene glycol + Mass of water

Mass of solution = 1.114 + 1.00 = 2.114 g

Now we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{2.114g}{1.070g/mL}=1.975mL

(a) Now we have to calculate the volume percent.

\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}\times 100=\frac{1mL}{1.975mL}\times 100=50.63\%

(b) Now we have to calculate the mass percent.

\text{Mass percent}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}\times 100=\frac{1.114g}{2.114g}\times 100=52.69\%

(c) Now we have to calculate the molarity.

\text{Molarity}=\frac{\text{Mass of ethylene glycol}\times 1000}{\text{Molar mass of ethylene glycol}\times \text{Volume of solution (in mL)}}

\text{Molarity}=\frac{1.114g\times 1000}{62.07g/mole\times 1.975L}=9.087mole/L

(d) Now we have to calculate the molality.

\text{Molality}=\frac{\text{Mass of ethylene glycol}\times 1000}{\text{Molar mass of ethylene glycol}\times \text{Mass of water (in g)}}

\text{Molality}=\frac{1.114g\times 1000}{62.07g/mole\times 1kg}=17.947mole/kg

(e) Now we have to calculate the mole fraction of ethylene glycol.

\text{Mole fraction of ethylene glycol}=\frac{\text{Moles of ethylene glycol}}{\text{Moles of ethylene glycol}+\text{Moles of water}}

\text{Moles of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Molar of ethylene glycol}}=\frac{1.114g}{62.07g/mole}=0.01795mole

\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar of water}}=\frac{1g}{18g/mole}=0.0555mole

\text{Mole fraction of ethylene glycol}=\frac{0.01795mole}{0.01795mole+0.0555mole}=0.244

6 0
2 years ago
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