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3 years ago

Which is generally true of enzymes in cells?

2 answers:

6 0

The answer is A.

Enzymes have extremely interesting properties that make them little chemical-reaction machines. The purpose of an enzyme in a cell is to allow the cell to carry out chemical reactions very quickly. These reactions allow the cell to build things or take things apart as needed. This is how a cell grows and reproduces.

Wewaii [24]3 years ago

4 0

The answer is A, enzymes allow reactions to proceed at body temperature.

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An unknown atom has 5

labwork [276]

Answer:

Explanation:

the atomic mass is away protons + neutrons, electrons are neglatable.

7 0

3 years ago

A sodium hydroxide solution is made by mixing 8.70 g NaOH with 100 g of water. The resulting solution has a density of 1.087 g/m

Ahat [919]

Answer:

Mass fraction = 0.08004

Mole fraction = 0.0377

Explanation:

Given, Mass of NaOH = 8.70 g

Mass of solution = 8.70 + 100 g = 108.70 g

$Mass\ fraction\ of\ NaOH=\frac {8.70}{108.70}$ = 0.08004

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{8.70\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 0.2175\ mol$

Given, Mass of water = 100 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{100\ g}{18.0153\ g/mol}$

$Moles\ of\ water= 5.5508\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ NaOH=\frac {n_{NaOH}}{n_{NaOH}+n_{water}}$

$Mole\ fraction\ of\ NaOH=\frac {0.2175}{0.2175+5.5508}=0.0377$

3 0

2 years ago

I cant seem to figure out ANY of these... help when you can pls :.)

Nonamiya [84]

The first one is some reaction with water even I am studying the same

7 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

$\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol$

and,

$\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0

3 years ago

What is the empirical formula of propene CH3 CH2 CH2

kiruha [24]

Answer:

CH

Explanation:

Molecular formula of propene : C₃H₆

Take the HCF of carbon and hydrogen atoms :

3 = 3
6 = 2 x 3

Then, we can write the formula as :

3CH
This means there are 3 moles present

Empirical Formula :

Molecular Formula / No. of moles
C₃H₆ / 3
CH

The empirical formula of propene is CH

6 0

1 year ago