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3 years ago

This number tells you the number of each type of atom in a compound.

Chemistry

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer:

avogadro's constant

Explanation:

this is the fixed number of the atoms in the molecule of an element

avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

that is all gases with same temperature and pressure will always have equal numbers of molecules

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The glycerol resulting from triacylglycerol metabolism can be converted to a form that can enter glycolysis. Identify the compou

marta [7]

Answer:

glycerol, pyruvate, glycerol -3-phosphate, dihydroxyacetone phosphate and glucose♡ hope this helps♡

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3 years ago

True or false, cilia and mucus help keep our lungs clean and healthy?

fiasKO [112]

Answer:

It does. They trap particles so they don’t enter.

Explanation:

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3 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

7 0

3 years ago

What is the mass of 2.5 mol of ca, which has a molar mass of g/mol

swat32

Atomic mass Ca = 40 a.m.u

1 mole Ca ----------- 40 g
2.5 mols Ca -------- ( mass Ca )

Mass Ca = 2.5 x 40 / 1

Mass Ca = 100 / 1

= 100 g of Ca

hope this helps!

3 0

3 years ago

What mass of water is produced by the combustion 1.2 x 10^26 molecules of butane ?

podryga [215]

Answer:

17.934 kg of water

Explanation:

If balanced equation is not given; this format can come in handy.

For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:

2CₙH₂ₙ₊₂ + (3n+1) O₂ → (2n)CO₂ + 2(n+1) H₂O

For butane:

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)

2 moles of butane gives 10 moles of water.

1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)

Mass of 1 mole of any substance is equal to it's molar mass

So, if 2 x N molecules of butane gives 10 x 18 g of water.

Then 1.2 x 10²⁶ molecules will give:

$\frac{1.2 \times 10^{26} \times 180}{2 \times 6.022 \times 10^{23}}$

= 17.934 x 10³ g of water

= 17.934 kg of water

5 0

3 years ago