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3 years ago

Calculate the mass of each element in the

Chemistry

1 answer:

Lelechka [254]3 years ago

7 0

Answer:

For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively

Explanation:

The computation of the mass of each element is given below:

As we know that

A1 mole of ammonium nitrate i.e. 2 mol N, 4 mol H, 3 mol

Now we multiply each of above by the molar masses

For N

= 14.0 g/mol × 2

= 28.0 gN

For H

= 1.0 g/mol × 4

= 4.0 gN

ANd, for O

= 16.0 g/mol × 3

= 48.0 gN

Hence, For Mass N, Mass H, and Mass O, the mass is 28.0 g N, 4.0 g H, and 48.0 g respectively

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Otra manera para llamar la aceleración negativa es_______

enyata [817]

Answer:

creo que es cinemática, eso creo

8 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

How many atoms are in 1.50 moles of Hg?

Oxana [17]

Answer:

Atoms=9.033*10^23 atoms

Explanation:

Atoms=no.of miles*Avogadro's no.

Atoms=1.5*6.022*10^23

Atoms=9.033*10^23 atoms

3 0

3 years ago

Read 2 more answers

What is the concentration (in M) of a sample of the unknown dye with an absorbance of 0.25 at 542 nm?

Sladkaya [172]

Answer:

see explanation below

Explanation:

Question is incomplete, so in picture 1, you have a sample of this question with the missing data.

Now, in general terms, the absorbance of a substance can be calculated using the beer's law which is the following:

A = εlc

Where:

ε: molar absortivity

l: distance of the light in solution

c: concentration of solution

However, in this case, we have a plot line and a equation for this plot, so all we have to do is replace the given data into the equation and solve for x, which is the concentration.

the equation according to the plot is:

A = 15200c - 0.018

So solving for C for an absorbance of 0.25 is:

0.25 = 15200c - 0.018

0.25 + 0.018 = 15200c

0.268 = 15200c

c = 0.268/15200

c = 1.76x10⁻⁵ M

5 0

3 years ago

Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338

Elden [556K]

Answer: The concentration of chloride ions in the solution obtained is 0.674 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KCl:

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

$0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol$

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

For magnesium chloride:

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

$0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol$

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = $(2\times 0.224)=0.448mol$

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

$\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M$

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

5 0

3 years ago