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3 years ago

Other than a base and a sugar unit, which of the following is a component of a nucleotide?

Chemistry

1 answer:

just olya [345]3 years ago

3 0

Answer:

C. phosphate group

Explanation:

took quiz on edg

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How is the way scientist use the word theory different from the way people use it in there everday language

weqwewe [10]

Answer:

Because Scientists Uses hypothesis and try tests and try proving their points while regular normal people use it to just state a statement

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3 years ago

All cells _____.

xeze [42]

The answer is the fourth choice, or Have membrane-bound organelles. Hope this helps you!

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3 years ago

Why is oxygen crucial to living things?

KATRIN_1 [288]

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3 years ago

when carbon is burned in the air, it reacts with oxygen to form carbon dioxide. when 22.8 g of carbon were burned in the presenc

OverLord2011 [107]

Answer:

So there is83.6g CO2 produced

Explanation:

Burning carbon with air has the following equation

C + O2 → CO2

For 1 mol Carbon, we have 1 mol O2 and 1 mol CO2

Step 2: Calculating moles

mole C = 22.8g / 12g/mole

Mole C = 1.9 mole

1.9 mole C will completely react

Since for each mole C there is 1 mole O2 and 1 mole CO2

This means there will also react 1.9 mole of 02, to be formed 1.9 mole of CO2

mole CO2 = mass CO2 / Molar mass CO2

mass CO2 = 1.9 mole CO2 * 44g/mole =<u>83.6g CO2</u>

In this reaction 18.2 g of O2 remained unreacted

we can control this: 79g - 18.2 g = 60.8g

1.9 mole * 32g/mol = 60.8g

So there is83.6g CO2 produced

4 0

3 years ago

Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu

pochemuha

Answer : The volume of $CO_2$ will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

$HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2$

First we have to calculate the mass of $HCO_3^-$ in tablet.

$\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g$

Now we have to calculate the moles of $HCO_3^-$ .

Molar mass of $HCO_3^-$ = 1 + 12 + 3(16) = 61 g/mole

$\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles$

Now we have to calculate the moles of $CO_2$ .

From the balanced chemical reaction, we conclude that

As, 1 mole of $HCO_3^-$ react to give 1 mole of $CO_2$

So, 0.0202 mole of $HCO_3^-$ react to give 0.0202 mole of $CO_2$

The moles of $CO_2$ = 0.0202 mole

Now we have to calculate the volume of $CO_2$ by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = $37^oC=273+37=310K$

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

$(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)$

$V=0.51411L=514.11ml$

Therefore, the volume of $CO_2$ will be, 514.11 ml

6 0

3 years ago