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3 years ago

How many atoms are present in 0.035 moles?

Chemistry

1 answer:

RUDIKE [14]3 years ago

7 0

Answer:

$2.11x10^{22}atoms$

Explanation:

Hello!

In this case, considering the Avogadro's number, which relates the number of particles and one mole, we can infer that 1 mole of any element contains 6.022x10²³ atoms as shown below:

$\frac{6.022x10^{23}atoms}{1mol}$

Thus, we compute the number of atoms in 0.035 moles as shown below:

$atoms=0.035mol*\frac{6.022x10^{23}atoms}{1mol} \\\\atoms=2.11x10^{22}atoms$

Best regards!

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ANSWER ASAP ITS CHEMISTRY HIGH SCHOOL 10 POINTS!

loris [4]

Answer:

3rd option. 1–butanamine

Explanation:

To name the compound above, the following must be observed:

1. Locate the functional group in the chain. In this case the functional group is amine.

2. Locate the longest continuous carbon chain. This gives the parent name of the compound. In this case, the longest chain has 4 carbon i.e butane.

3. Since the functional group is amine, the parent name becomes butanamine i.e replacing the –e at the end in butane with –amine

4. Indicate the position of the functional group in the chain. In this case the functional group is at carbon 1

5. Name the compound by putting the above together.

The name of the compound is:

1–butanamine or butan–1–amine

5 0

3 years ago

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

8 0

3 years ago

Dl water will test<br> _________for protein?

Ann [662]

Answer:

Total protein range. The normal range for total protein is between 6 and 8.3 grams per deciliter (g/dL). This range may vary slightly among laboratories.

Explanation:

5 0

3 years ago

In an experiment, students were given an unknown mineral. The unknown mineral was placed in 150 ml of water. Once in the water,

Dahasolnce [82]

Answer:

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

But from the question

volume = final volume of water - initial volume of water

volume = 165 - 150 = 15 mL

We have

$density = \frac{225}{15} = 15 \\$

We have the final answer as

Hope this helps you

5 0

3 years ago

if a sample of gas at 25.2 c has a volume of 536mL at 637 torr, what will its volume be if the pressure is increased to 712 torr

nignag [31]

Considering ideal gas:
PV= RTn

T= 25.2°C = 298.2 K

P1= 637 torr = 0.8382 atm

V1= 536 mL = 0.536 L

:. R=0.082 atm.L/K.mol

:. n= (P1V1)/(RT) = ((0.8382 atm) x (0.536 L))/
((0.082 atmL/Kmol) x (298.2K))

:. n= O.0184 mol

Then,
P2= 712 torr = 0.936842 atm

V2 = RTn/P2 = [(0.082atmL/
Kmol) x (298.2K) x (0.0184mol) ]/(0.936842atm)

:.V2 = 0.4796 L
OR
V2 = 479.6 ml

6 0

3 years ago