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Likurg_2 [28]
3 years ago
5

The rate of effusion of a particular gas was measured and found to be 24.0 mL/min. Under the same conditions, the rate of effusi

on of pure methane (CH4) gas is 47.8 mL/min. What is the molar mass of the unknown gas?

Chemistry
1 answer:
Norma-Jean [14]3 years ago
3 0

Answer:

63.6g/mol

Explanation:

Use the equation of effusion:

47.8/24=\sqrt{x/16.04}

solve for x you get 63.6g/mol

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calculate the mass required to prepare 2.5 L of 1.0 M NaOH solution. Given that the molar mass for NaOH is 40 g/mol.
Helen [10]

Answer:

The required mass to prepare 2.5 L of 1.0 M NaOH solution is 100 g

Explanation:

We do this by preparing the equation:

Mass = concentration (mol/L) x volume (L) x Molar mass

Mass = 1.0 M x 2.5 L x 40 g/mol

Mass = 100 g

3 0
2 years ago
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What are five kinds of evidence you can use to determine if a chemical reaction has occurred?
telo118 [61]
Change of odor, change of color, change of temperature, change of energy, or loss of heat
5 0
2 years ago
For the reaction NH4Cl (s)→NH3 (g)  + HCl (g) at 25°C, ΔH = 176 kJ/moland ΔS = 0.285 kJ/(mol - K).
rjkz [21]

Answer:

91kj/mol;no

Explanation:

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3 0
2 years ago
What mass of AgCl is produced when 53.42 g of AgNO3 reacts with 14.19 g of NaCl?
Leya [2.2K]

Answer is: mass of silver chloride is 34,82 g.

Chemical reaction: AgNO₃ + NaCl → AgCl + NaNO₃.

<span>m(AgNO</span>₃) = 53,42 g..<span>
m(NaCl) = 14,19 g.
n(AgNO</span>₃) = m(AgNO₃) ÷ M(AgNO₃).<span>
n(AgNO</span>₃) = 53,42 g ÷ 169,87 g/mol.

n(AgNO₃) = 0,314 mol.

n(NaCl) = 14,19 g ÷ 58,4 g/mol.

n(NaCl) = 0,242 mol; limiting reactant.

From chemical reaction: n(NaCl) : n(AgCl) = 1 : 1.<span>
n</span>(AgCl)<span> = 0,242 mol.
m</span>(AgCl) = 0,252 mol · 143,32 g/mol.

m(AgCl) = 34,82 g.

7 0
3 years ago
What volume will 2.0 moles of nitrogen occupy at .947 atm and 20° C ? ____ L
r-ruslan [8.4K]

Answer:

The volume is 50, 74 liters

Explanation:

We use the formula PV = nRT. The temperature in Kelvin is = 273+ 20 =293K

PV=nRT ---> V = (nRT) / P

V= (2.0 mol x 0,082 l atm /K mol x 293 K) / 0,947 atm

V=50,74128828 l

4 0
2 years ago
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