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2 years ago

List three differences between liquids and gases

Chemistry

1 answer:

Annette [7]2 years ago

3 0

Answer:

Gas:

- No fixed shape or volume

- Molecules are very loosely packed

- Flows in all the directions

Liquid:

- No fixed shape but has volume

- Molecules are closely packed

- Always flows from higher to lower level

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4. One gram of of ............... contains one type of atoms. a. iron b. ammonia c. hydrogen chloride d. water

patriot [66]

Answer: hydrogen chloride.

Explanation:

7 0

2 years ago

A student weighed out a 2.055 g sample of a cobalt chloride hydrate, ConClmpH2O, where n, m, and p are integer values to be dete

topjm [15]

The given mass of cobalt chloride hydrate = 2.055 g

A sample of cobalt chloride hydrate was heated to drive off waters of hydration and the anhydrate was weighed.

The mass of anhydrous cobalt chloride = 1.121 g anhydrate.

The mass of water lost during heating = 2.055 g - 1.121 g = 0.934 g

Converting mass of water of hydration present in the hydrate to moles using molar mass:

Mass of water = 0.934 g

Molar mass of water = 18.0 g/mol

Moles of water = $0.934 g * \frac{1 molH_{2}O }{18 g H_{2}O } =0.0519 mol H_{2}O$

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3 years ago

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.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

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

$Q_{metal} = Q_{water}$

Where:

$Q_{water}$ - 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:

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