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2 years ago

Calculate the molecular mass of the element

Chemistry

1 answer:

Elanso [62]2 years ago

4 0

Answer:

Cu3 (PO4)2

3×64 + 2×(31+ 4×16)

192 + 2×(31+64)

192 + 2×(95)

192+190

=382g

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Will reward brainliest<br> Explain why all arrhenius acids can be considered bronsted lowry acids

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Answer:

Explanation:

A bronsted lowry acid just means that it donates a proton.

An arrhenius acid dissolves in water to donate a proton

the only difference is that an arrhenius acid must dissolve in water but it still donates a proton so it is considered a bronsted lowry acid

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2 years ago

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3 years ago

A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a

KatRina [158]

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

Partial pressure of methane (pCH₄): 431 mmHg

Partial pressure of carbon dioxide (pCO₂): 504 mmHg

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

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XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

3 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