Answer:
3000 kJ/kg
Explanation:
The calorific value of a substance is the amount of heat produced per unit mass by the combustion of the substance.
It is given by:
where
Q is the amount of heat released
m is the mass of the fuel
In this problem, we have:
m = 60 kg is the mass of fuel
is the amount of heat released
Therefore, the calorific value of the fuel is:
Answer:
59.92 × 10²³ atoms are in 9.95 moles of iron
1.8 ×10²² molecules are in 0.03 moles of Carbon dioxide
1.19 moles are found in 7.20 x 10^23 atoms of platinum
Answer: A) Inconclusive; you would not know which of the two variables caused the change.
Explanation:
When you set up an experiment, you must make sure that you control the variables such that only one independent variable changes at a time, while all the remainder conditions (the other independent variables) are controlled (fixed).
By observing (measuring) the dependent variable, while only one independent variable changes you can understandhow such independent variable explains (determines) the dependent variable, leading to a conclusion.
Conversely, if two or more independent variables change at a time, then there is no way that you can tell how the output (dependent variable) is related with one or other of the changes of the indipendent variables. You wolud not be able to discriminate (distinguish) the effect of one or other variable, making the experiment inconclusive
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Answer:
the result for the following are (a) P is directly proportional to n
(b) V is directly proportional to T (c) P is directly proportional to T (d) T is inversly proportional to V (e) P is inversely proportional to V
Answer:
0.42%
Explanation:
<em>∵ pH = - log[H⁺].</em>
2.72 = - log[H⁺]
∴ [H⁺] = 1.905 x 10⁻³.
<em>∵ [H⁺] = √Ka.C</em>
∴ [H⁺]² = Ka.C
∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.
<em>∵ Ka = α²C.</em>
Where, α is the degree of dissociation.
<em>∴ α = √(Ka/C) </em>= √(8.065 x 10⁻⁶/0.45) = <em>4.234 x 10⁻³.</em>
<em>∴ percentage ionization of the acid = α x 100</em> = (4.233 x 10⁻³)(100) = <em>0.4233% ≅ 0.42%.</em>
