All categories

3 years ago

40 Points!!! Match the correct definition with the correct keyword.

Chemistry

1 answer:

My name is Ann [436]3 years ago

8 0

Answer:

Okay the Answers are top to bottom.

5, 4, 1, 3, 2

Explanation:

You might be interested in

If the male monster has a straight tail, what does his genotype have to be?

Anarel [89]

i think a is the answer

8 0

3 years ago

A gas has a volume of 6.35 L at 88.6 kPa. What will be the volume at standard pressure? Standard Pressure = 101.3 kPa

alukav5142 [94]

Answer:

5.55 L

Explanation:

This excersise can be solved by the Boyle's law.

This law for gases states that the pressure of a gas in a vessel is inversely proportional to the volume of the vessel.

P₁ . V₁ = P₂ . V₂

The law comes from the Ideal Gases Law, in the first term.

P . V = n . R . T In this case, n . R . T are all constant.

6.35 L . 88.6 kPa = 101.3 kPa . V₂

V₂ = (6.35 L . 88.6 kPa) / 101.3 kPa

V₂ = 5.55 L

It is inversely proportional because, as it happened in this case, pressure was increased, therefore volume decreased.

5 0

3 years ago

Which example best describes a reflex action?

Pepsi [2]

I think the answer would be b because your throat is reacting to the cough?

6 0

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:

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

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

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

5 0

3 years ago

The least polar bond is found in a molecule of hi hf hcl hbr

mario62 [17]

Answer:

HI is least polar

Explanation:

5 0

3 years ago