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3 years ago

A quantity of gas weighing 7.1 g at 741 torr and 44 celius occupies 5.41 L. Find its molar mass.

1 answer:

3 0

1 atmosphere = 760 torr
R = 0.082
44 degrees Celsius = 317 degrees Kelvin

$PV=NRT\\ \frac{744}{760}(5.41)=N(0.082)(317)\\ \frac{4025.04}{760}=25.994N\\\frac{4025.04}{19755.44}=N\\N=\frac{50313}{246943}\\0.203743$

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8.38e-21 Q^2 -1.07e-23 Q^2 +3.15e+19 Solve for Q

GarryVolchara [31]

8.38e -21Q^2 -1.07e -23Q^2 +3.15e +19
= 10.46e -44Q^2 + 19
44Q^2 = 10.46e + 19
Q^2 = 523/2200e + 19/44
Q1= ≈ -1.03828
Q2= ≈ 1.03828

4 0

4 years ago

If two electrons in the same atom have the same value of "l"

lesya692 [45]

Answer:

If two electrons in the same atom have the same value of "l", they are If two electrons in the same atom have the same value of "l", they are in the same orbital. in the same sublevel, but not necessarily in the same level. in different levels and in different shaped orbitals. in the same level, but different sublevel.

3 0

3 years ago

In the laboratory a student finds that it takes 44.0 Joules to increase the temperature of 10.6 grams of solid zinc from 24.9 to

BaLLatris [955]

Answer:

The specific heat of zinc is 0.361 J/g°C

Explanation:

Step 1: Data given

44.0 J needed

Mass of solid zinc = 10.6 grams

Initial temperature = 24.9 °C

Final temperature = 36.4 °C

Step 2: Calculate the specific heat of zinc

Q = m*c*ΔT

⇒ with Q = heat (in Joule) = 44.0 J

⇒ with m = the mass of the solid zinc = 10.6 grams

⇒ with c = the specific heat of the zinc = TO BE DETERMINED

⇒ with ΔT = The change in temperature = T2-T1 = 36.4 °C - 24.9 °C = 11.5 °C

44.0 J = 10.6 grams * c * 11.5°C

c = 44.0 J / (10.6g * 11.5 °C)

c = 0.361 J/g°C

The specific heat of zinc is 0.361 J/g°C

7 0

4 years ago

Name the reaction & mechanism of the following reactions:

gtnhenbr [62]

Answer:

double displacement reaction

6 0

3 years ago

Read 2 more answers

2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom

lakkis [162]

Answer: The average atomic mass of element Zinc is 65.40 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For $_{30}^{64}\textrm{Zn}$ isotope:

Mass of $_{30}^{64}\textrm{Zn}$ isotope = 63.929 amu

Percentage abundance of $_{30}^{64}\textrm{Zn}$ isotope = 48.63 %

Fractional abundance of $_{30}^{64}\textrm{Zn}$ isotope = 0.4863

For $_{30}^{66}\textrm{Zn}$ isotope:

Mass of $_{30}^{66}\textrm{Zn}$ isotope = 65.926 amu

Percentage abundance of $_{30}^{66}\textrm{Zn}$ isotope = 27.90 %

Fractional abundance of $_{30}^{66}\textrm{Zn}$ isotope = 0.2790

For $_{30}^{67}\textrm{Zn}$ isotope:

Mass of $_{30}^{67}\textrm{Zn}$ isotope = 66.927 amu

Percentage abundance of $_{30}^{67}\textrm{Zn}$ isotope = 4.10 %

Fractional abundance of $_{30}^{67}\textrm{Zn}$ isotope = 0.0410

For $_{30}^{68}\textrm{Zn}$ isotope:

Mass of $_{30}^{68}\textrm{Zn}$ isotope = 67.925 amu

Percentage abundance of $_{30}^{68}\textrm{Zn}$ isotope = 18.75 %

Fractional abundance of $_{30}^{68}\textrm{Zn}$ isotope = 0.1875

For $_{30}^{70}\textrm{Zn}$ isotope:

Mass of $_{30}^{70}\textrm{Zn}$ isotope = 69.925 amu

Percentage abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.62 %

Fractional abundance of $_{30}^{70}\textrm{Zn}$ isotope = 0.0062

Putting values in equation 1, we get:

$\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]$

$\text{Average atomic mass of Zinc}=65.40amu$

Hence, the average atomic mass of element Zinc is 65.40 amu.

7 0

4 years ago