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zloy xaker [14]
3 years ago
14

A room with dimensions 7.00m×8.00m×2.50m is to be filled with pure oxygen at 22.0∘c and 1.00 atm. The molar mass of oxygen is 32

.0 g/mol.
Chemistry
1 answer:
patriot [66]3 years ago
3 0

Answer:

Mass of O₂ present in the room: 185200 g

185.2 kg

Quantity of O₂ = 5787.5

Explanation:

The ideal gases Law can solved, what is ths mass of O₂ present in the room, or the quantity of O₂ (moles), present in the room.

P . V = n . R . T

P = 1 atm

V = volume of the room

n = moles

R = 0.082 L.atm/mol.K

T = T° in K

T° C + 273 → 22°C + 273 = 295K

Let's find out the volume of the room

7 m . 8 m . 2.5 m = 140 m³

Units of R, are in L, so we must convert m³ to dm³

1dm³ = 1 L

140m³ . 1000 = 140000 dm³

1.4x10⁵L

1 atm . 1.4x10⁵L = n . 0.082 L.atm/mol.K . 295K

(1 atm . 1.4x10⁵L) / (0.082 L.atm/mol.K . 295K)  = n

5787.5 moles = n

molar mass . moles = mass

5787.5 moles O₂ . 32 g/m = 185200 g of O₂

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nexus9112 [7]

Answer:

2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.

Explanation:

  • For balancing a chemical equation, we should apply the law of conversation of mass. It states that the no. of atoms in the reactants side is equal to that of the products side.

So, the balanced equation:

<em>2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.</em>

It is clear that 2.0 moles of C₃H₇BO₃ is completely burned in 8 m oles of oxygen and produce 6 moles of CO₂, 7 moles of H₂O and 1 mole of B₂O₃.

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The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

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K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

Concentration = 0.35 M

HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

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