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3 years ago

Please answer question #4

Chemistry

2 answers:

mixer [17]3 years ago

8 0

Answer:

force = 1500N
work done = 15000 joules

Explanation:

Force = mass × acceleration

Force = 500 × 3 = 1500 Newtons

work done = force × displacement

Work done = 1500 × 10 = 15000j = 15kj

( j = joule, kj = kilo joules )

trasher [3.6K]3 years ago

7 0

Answer:

F=500×3 = 1500 N

W = 1500×10 =15000 Nm

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Calculate the mass of CO2 that can be produced if the reaction of 54.0 g of propane and sufficient oxygen has a 64.0% yield.

Oduvanchick [21]

Answer:

103.9 g

Explanation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

First we convert 54.0 g of propane (C₃H₈) into moles, using its molar mass:

54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈

Then we convert 1.23 moles of C₃H₈ into moles of CO₂, using the stoichiometric coefficients:

1.23 mol C₃H₈ * $\frac{3molCO_2}{1molC_3H_8}$ = 3.69 mol CO₂

We convert 3.69 moles of CO₂ into grams, using its molar mass:

3.69 mol CO₂ * 44 g/mol = 162.36 g

And apply the given yield:

162.36 g * 64.0/100 = 103.9 g

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3 years ago

Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002

marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The $pK_a=4.756$

The value of the dissociation constant = $K_a$

$pK_a=-\log[K_a]$

$K_a=10^{-4.756}=1.754\times 10^{-5}$

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

$HAc\rightleftharpoons H^++Ac^-$

Initially

At equilibrium ;

(c-cα) cα cα

The expression of dissociation constant is given as:

$K_a=\frac{[H^+][Ac^-]}{[HAc]}$

$1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}$

$1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}$

$1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}$

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

$[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M$

The pH of the solution ;

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

$=-\log[0.0001901 M]=3.72$

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3 years ago

Substance X has a fixed volume, and the attraction between its particles is strong. Substance Y has widely spread out particles

DedPeter [7]

Answer: X is a Solid; Y is a Gas

Explanation:

There are three (3) states of matter. They are: Solid, Liquid and Gases.

Substance X and Y, belong to the states of matter.

A Solid is a substance that retains its SIZE and SHAPE without need of a container (as opposed to a liquid or gas).

Thus, it will most likely be concluded that: substance X is a Solid; while Y is a Gas

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3 years ago

How many grams of dry nh4cl need to be added to 2.50 l of a 0.800 m solution of ammonia, nh3, to prepare a buffer solution that

tia_tia [17]

Answer:

The correct answer is 574.59 grams.

Explanation:

Based on the given information, the number of moles of NH₃ will be,

= 2.50 L × 0.800 mol/L

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pH + pOH = 14.00

pOH = 14.00 - pH

pOH = 14.00 - 8.53

pOH = 5.47

The Kb of ammonia given is 1.8 * 10^-5. Now pKb = -logKb,

= -log (1.8 ×10⁻⁵)

= 5.00 - log 1.8

= 5.00 - 0.26

= 4.74

Based on Henderson equation:

pOH = pKb + log ([salt]/[base])

pOH = pKb + [NH₄⁺]/[NH₃]

5.47 = 4.74 + log ([NH₄⁺]/[NH₃])

log([NH₄⁺]/[NH₃]) = 5.47-4.74 = 0.73

[NH₄⁺]/[NH₃] = 10^0.73= 5.37

[NH₄⁺ = 5.37 × 2 mol = 10.74 mol

Now the mass of dry ammonium chloride required is,

mass of NH₄Cl = 10.74 mol × 53.5 g/mol

= 574.59 grams.

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