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2 years ago

What is the molar Mass of 2.4g of oxygen

1 answer:

4 0

<span>2.4g of a compound of carbon, hydrogen and oxygen gave on combustion, 3.52g of CO2 and 1.44g of H2O. The relative molecular mass of the compound was found to be 60. a)What are the masses of carbon, hydrogen and oxygen in 2.4g of the compound? b)What are the emperical and ..</span>

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Part I: Multiple Choice Questlons

mina [271]

Answer:

A cuz a heterogeneous mixture is no uniform

7 0

2 years ago

Read 2 more answers

The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.

madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

$H_3BO_3\rightarrow H^+H_2BO_3^-$

at t=0 cM 0 0

at eqm $c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 4.4 M and $\alpha$ = ?

$K_a=5.8\times 10^{-10}$

Putting in the values we get:

$5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}$

$(\alpha)=0.000011$

$[H^+]=c\times \alpha$

$[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M$

Also $pH=-log[H^+]$

$pH=-log[4.8\times 10^{-5}]=4.3$

Thus pH of a 4.4 M $H_3BO_3$ solution is 4.3

3 0

3 years ago

When 8.0 grams of sodium hydroxide is dissolved in sufficient water to make 400. mL of solution, what is the concentration of th

nata0808 [166]

Answer:

The concentration of this sodiumhydroxide solutions is 0.50 M

Explanation:

Step 1: Data given

Mass of sodium hydroxide (NaOh) = 8.0 grams

Molar mass of sodium hydroxide = 40.0 g/mol

Volume water = 400 mL = 0.400 L

Step 2: Calculate moles NaOH

Moles NaOH = mass NaOH / molar mass NaOH

Moles NaOH = 8.0 grams / 40.0 g/mol

Moles NaOh = 0.20 moles

Step 3: Calculate concentration of the solution

Concentration solution = moles NaOH / volume water

Concentration solution = 0.20 moles / 0.400 L

Concentration solution = 0.50 M

The concentration of this sodiumhydroxide solutions is 0.50 M

6 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

meriva

Answer:

The standard enthalpy change for the reaction at $25^{0}\textrm{C}$ is -2043.999kJ

Explanation:

Standard enthalpy change ( $\Delta H_{rxn}^{0}$ ) for the given reaction is expressed as:

$\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$