Given the balanced equation representing a reaction:

In a room full of air, the air is mainly composed of Nitrogen and Oxygen molecules (both at room temperature). Find (to two sign

zloy xaker [14]

Explanation:

Expression for the $v_{rms}$ speed is as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

where, $v_{rms}$ = root mean square speed

k = Boltzmann constant

T = temperature

M = molecular mass

As the molecular weight of oxygen is 0.031 kg/mol and R = 8.314 J/mol K. Hence, we will calculate the value of $v_{rms}$ as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

= $\sqrt{\frac{3 \times 8.314 J/mol K \times 309.02 K}{0.031 kg/mol}}$

= 498.5 m/s

Hence, $v_{rms}$ for oxygen atom is 498.5 m/s.

For nitrogen atom, the molecular weight is 0.028 kg/mol. Hence, we will calculate its $v_{rms}$ speed as follows.

$v_{rms} = \sqrt{\frac{3kT}{M}}$

= $\sqrt{\frac{3 \times 8.314 J/mol K \times 309.92 K}{0.028 kg/mol}}$

= 524.5 m/s

Therefore, $v_{rms}$ speed for nitrogen is 524.5 m/s.

3 0

3 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago

How much thermal energy (q) is required to heat 15.2g of a metal (specific heat=0.397J/C) from 21.0C to 40.3C? Show steps please

Elena L [17]

Q = ?

Cp = 0.397 J/ºC

Δt = 40.3 - 21.0 => 19.3 ºC

m = 15.2 g

Q = m x Cp x Δt

Q = 15.2 x 0.397 x 19.3

Q ≈ 116.46 J

hope this helps!

7 0

4 years ago

What is the volume of 55 grams of carbon dioxide at STP

e-lub [12.9K]

Answer:

A. 2.45 L - Answer my answers is right.

7 0

3 years ago

Consider the reaction. mc026-1.jpg At equilibrium at 600 K, the concentrations are as follows. [HF] = 5.82 x 10-2 M [H2] = 8.4 x

nadezda [96]

Answer:

2.1 x 10⁻².

Explanation:

Generally, the equilibrium constant (Keq) is the product of the concentration of the reaction products divided by the product of the concentration of the reaction reactants, each term is raised to a power equal to its coefficient in the balanced chemical reaction.

For the given reaction: 2HF → H₂ + F₂,

Keq = [H₂][F₂] / [HF]²,

Keq = [H₂][F₂] / [HF]² = [8.4 x 10⁻³][8.4 x 10⁻³] / [5.82 x 10⁻²]² = 2.1 x 10⁻².

8 0

3 years ago