All categories

2 years ago

Determine the mass in grams of 125 mol of neon

Chemistry

1 answer:

Vesnalui [34]2 years ago

5 0

Molar mass of Neon ( Ne ) = 20.1797 g/mol

m = n * mm

m = 125 * 20.1797

m = 2522.4625 g

hope this helps!

You might be interested in

Which of the following substances found in tobacco smoke stimulates the brain? A. acetone B. toluene C. nicotine D. ammonia Plea

Vaselesa [24]

The answer is C. Nicotine is the substance found in tobacco smoke that stimulates the brain.

7 0

3 years ago

Read 2 more answers

What is one advantage that proteins derived from beans have over proteins derived from red meats

lisov135 [29]

Answer: B. Proteins derived from beans have less saturated fat than proteins derived from red meats.

3 0

2 years ago

Balance the following chemical equation by providing the correct coefficients HBr+Ba(OH)2 ———— BaBr2+H2O

stealth61 [152]

Answer:

2HBr + Ba(OH)2 = BaBr2 + 2H20

4 0

2 years ago

Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, $C_{Ag}$ = 78 wt%

concentration of Pd, $C_P_d$ = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, $A_A_g$ = 107.87 g/mol

atomic weight of iron, $A_P_d$ = 106.4 g/mol

Step 1: determine the average density of the alloy

$\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }$

$\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3$

Step 2: determine the average atomic weight of the alloy

$A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }$

$A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole$

Step 3: determine unit cell volume

$V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}$

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

$V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell$

Step 4: determine the unit cell edge length

Vc = a³

$a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm$ = 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0

3 years ago

A student is investigating the effect of concentration on the colour of a solution of copper sulfate.She wished to make up 250cm

IrinaK [193]

Answer:

79.8g/dm³

Explanation:

As you can see, the solution in the problem contains 0.5 moles of copper sulfate per dm³. To solve this question we must convert these moles to grams using its molar mass (Molar mass CuSO4 = 159.609g/mol) as follows:

0.5mol CuSO4/dm³ * (159.609g/mol) =

3 0

2 years ago