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mina [271]
3 years ago
15

The atomic mass of hydrogen-1 is 1.008 and the atomic mass of helium-4 is 4.003. Find the difference in grams between 4 moles of

hydrogen and one mole of helium.
A) 4.81 x10-26 g

B) 0.029 g

C) 1.7 x 1022 g

D) 0 grams
Chemistry
2 answers:
CaHeK987 [17]3 years ago
7 0

4 moles of hydrogen  = 4 * 1.008 = 4.032 grams

1 mole of helium = 4.003 grams

Difference is  4.032 - 4.003

= 0.029 g

melomori [17]3 years ago
4 0

Answer:

The correct answer is option B.

Explanation:

The atomic mass of hydrogen-1 = 1.008 g/mol

The atomic mass of helium-4 = 4.003 g/mol

Mass of 4 mole of hydrogen = 4 × 1.008 g/mol = 4.032 g

Mass of 1 mole of helium = 1 × 4.003 g/mol = 4.003 g

Difference of between 4 moles of hydrogen and 1 moles of helium :

4.032 g - 4.003 g = 0.029 g

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Answer:

kinetic

Explanation:

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3 years ago
How much faster does ethane (C2H6) gas travel compared to chlorine gas?
zimovet [89]

Answer:

\large \boxed{\text{54 $\, \%$ faster }}

Explanation:

v_{\text{rms}} \propto \sqrt{\dfrac{3RT}{M}

if temperature is constant.  

v_{\text{rms}} \propto \sqrt{\dfrac{1}{M}

if we are comparing two gases,

\dfrac{v_{2}}{v_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}

Let chlorine be Gas 1 and ethane be Gas 2

Data:

M₁ =  70.91 g/mol

M₂ = 30.07 g/mol

Calculation

\begin{array}{rcl}\dfrac{v_{2}}{v_{1}} & = & \sqrt{\dfrac{M_{1}}{M_{2}}}\\\\& = & \sqrt{\dfrac{70.91}{30.07}}\\\\& = & \sqrt{2.358}\\\\& = & \mathbf{1.54}\\\end{array}\\\text{Ethane molecules travel at 1.54 times the speed of chlorine molecules}\\\text{or $\large \boxed{\textbf{54 $\%$ faster }}$ than chlorine molecules}

5 0
3 years ago
Name the following compounds:
ankoles [38]
A) cesium chloride
B) barium oxide
C) potassium sulfide
D) beryllium chloride
E) hydrogen bromide
F) aluminum fluoride
3 0
3 years ago
A space air is at a temperature of 75 oF, and the relative humidity (RH) is 45%. Using calculations, find: (a) the partial press
earnstyle [38]

Answer:

A) Partial Pressure of dry air = 13.32 KPa

Partial Pressure of water vapour = 1.332 KPa

B) Humidity ratio; X = 0.0691

C) V_p = 0.8384 m³/Kg

Explanation:

A) We are given;

Temperature = 75°F

Relative Humidity = 45%

Now,to calculate the partial pressure, we will use the relationship;

Relative Humidity = (Partial Pressure/Vapour Pressure) × 100%

Making partial pressure the subject;

Partial Pressure = Relative Humidity × Vapour Pressure/100%

From the first table attached, at temperature of 75°F, the vapor pressure is 29.6 × 10^(-3) bar = 29.6 KPa

Thus;

Partial Pressure of dry air = (45 × 29.6)/100

Partial Pressure of dry air = 13.32 KPa

From online values, vapour pressure of water vapour at 75°F = 2.96 KPa

Thus;

Partial Pressure of water vapour = (45 × 2.96)/100 = 1.332 KPa

B) humidity ratio of moist air is given as;

X = 0.62198 pw / (pa - pw)  

where;

pw = partial pressure of the water vapor in moist air

pa = atmospheric pressure of the moist air

Thus;

X = (0.62198 × 1.332)/(13.32 - 1.332)  

X = 0.0691

C) Formula for moist air specific volume is;

V_p = (1 + (xRw/Ra) × RaT/p

Where;

V_p is specific volume

T is temperature = 75°F = 297.039 K

p is barometric pressure which in this case is standard sea level pressure = 101.325 KPa

pw is partial pressure of the water vapor in moist air = 1.332 KPa

Rw is individual gas constant for water = 0.4614 KJ/Kg.K

Ra is individual gas constant for air = 0.2869 KJ/Kg.K

V_p = (1 + (0.0691 * 0.4614/0.2869)) × 0.286.9 * 297.039/101.325

V_p = 0.8384 m³/Kg

6 0
3 years ago
A 0.530 M Ca(OH)2 solution was prepared by dissolving 36.0 grams of Ca(OH)2 in enough water. What is the total volume of the sol
Gekata [30.6K]

Answer:

0.917

Explanation:

3 0
3 years ago
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