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3 years ago

How many molecules are in 1.2 moles of H20?

Chemistry

1 answer:

Otrada [13]3 years ago

4 0

<h3>Answer:</h3>

7.226 × 10^23 molecules.

<h3>Explanation:</h3>

A compound is a substance that is made by two or more atoms from different elements.

A mole of a compound contains a number of molecules equivalent to Avogadro's number, 6.022 × 10^23.
That is, one mole of a compound contains 6.022 × 10^23 molecules.

In this case we are given;

Number of moles of H₂O as 1.2 moles

But, 1 mole of H₂O contains 6.022 × 10^23 molecules.

We are required to calculate the number of molecules present;

To calculate the number of molecules we are going to multiply the number of molecules in one mole by the number of moles.

Therefore;

Number of molecules = 1.2 moles × 6.022 × 10^23 molecules/mole

= 7.226 × 10^23 molecules.

Thus, 1.2 moles of water contains 7.226 × 10^23 molecules.

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A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0

miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

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$Q_{metal} - Q_{water} = 0$

$Q_{metal} = Q_{water}$

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$Q_{water}$ - Heat received by water, measured in joules.

$Q_{metal}$ - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

$m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})$

The specific heat of the metal is cleared within equation above:

$c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}$

If we know that $m_{water} = 0.237\,kg$ , $m_{metal} = 0.244\,kg$ , $c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{w,o} = 10\,^{\circ}C$ , $T_{m,o} = 100.10\,^{\circ}C$ and $T = 15.30\,^{\circ}C$ , the specific heat of molybdenum is:

$c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}$

$c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}$

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

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Answer:

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Explanation:

Balancing:

Cu + 4HNO3 ---> Cu(NO3)2 + 2 NO2 + 2H2O.

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How many molecules are in 1.2 moles of H20?​

How many molecules are in 1.2 moles of H20?