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3 years ago

how much heat energy is needed to raise the temperature of 78.4g of aluminum from 19.4 degrees c to 98.6 degrees c.

Chemistry

1 answer:

Iteru [2.4K]3 years ago

7 0

The heat that is needed to raise the temperature of 78.4 g of aluminium from 19.4 °c to 98.6°c is 5600.77 j

calculation

Heat(Q) = mass(M) x specific heat capacity (C) x change in temperature(ΔT)

where;

Q=?

M = 78. 4 g

C=0.902 j/g/c

ΔT=98.6°c -19.4°c =79.2°c

Q is therefore = 78.4 g x 0.902 j/g/c x 79.2°c =5600.77 j

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How many moles are in 3.57 x 10 32 particles Ca?

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Answer:

$\boxed {\boxed {\sf 592,826,303.6 \ moles \ of \ Ca }}$

Explanation:

To convert from moles to particles, we must Avogadro's Number.

$6.022*10^{23}$

This number tells us the number of particles (atoms, ions, molecules, etc.) in 1 mole of a substance. In this case, it is the particles of calcium in 1 mole of calcium.

$6.022*10^{23} \ particles \ Ca / mole \ Ca$

We can use Avogadro's Number as a ratio or fraction.

$\frac{6.022 *10^{23} \ particles \ Ca}{1 \ mol \ Ca}$

Multiply this by the given number of particles (3.57*10³²).

$3.57*10^{32} \ particles \ Ca*\frac{6.022 *10^{23} \ particles \ Ca}{1 \ mol \ Ca}$

Flip the fraction so the particles of calcium will cancel out.

$3.57*10^{32} \ particles \ Ca*\frac{1 \ mol \ Ca}{6.022 *10^{23} \ particles \ Ca}$

$3.57*10^{32} *\frac{1 \ mol \ Ca}{6.022 *10^{23} }$

$\frac{3.57*10^{32} \ mol \ Ca}{6.022 *10^{23} }$

$592826303.6 \ mol \ Ca$

There are 592,826,303.6 moles of calcium in 3.57*10³² particles of calcium.

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Answer:

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