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2 years ago

6

When 0.2 kg of gasoline undergoes combustion, 18500 kJ of energy are released. Express the heat released, 18500 kJ in calories.

(Given 4.184 J = 1 cal)

1 answer:

mezya [45]2 years ago

5 0

The problem above is simply asking to convert one unit of energy to another of unit energy. We convert measurements by multiplying conversion factors. The conversion factor, for this case, is 1 cal / 4.184 kJ. We multiply this value to the measurement.

18500 kJ (1/4.184) = 4421.61 calories

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2 years ago

2 NO + O22 NO2 is second order in NO and first order in O2. Complete the rate law for this reaction in the box below. Use the fo

riadik2000 [5.3K]

Answer : The value of rate of reaction is $1.35\times 10^{-8}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The given chemical equation is:

$2NO+O_2\rightarrow 2NO_2$

Rate law expression for the reaction is:

$\text{Rate}=k[NO]^a[O_2]^b$

As per question,

a = order with respect to $NO$ = 2

b = order with respect to $O_2$ = 1

Thus, the rate law becomes:

$\text{Rate}=k[NO]^2[O_2]^1$

Now, calculating the value of rate of reaction by using the rate law expression.

Given :

k = rate constant = $9.87\times 10^3M^{-2}s^{-1}$

[NO] = concentration of NO = $7.86\times 10^{-3}M$

$[O_2]$ = concentration of $O_2$ = $2.21\times 10^{-3}M$

Now put all the given values in the above expression, we get:

$\text{Rate}=(9.87\times 10^3M^{-2}s^{-1})\times (7.86\times 10^{-3}M)^2\times (2.21\times 10^{-3}M)^1$

$\text{Rate}=1.35\times 10^{-8}Ms^{-1}$

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2 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

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Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

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2 years ago