All categories

3 years ago

Water and minerals move up the xylem because of a vacuum created by _____.

Chemistry

1 answer:

Galina-37 [17]3 years ago

4 0

Answer:

The right answer is B) evaporation

Explanation:

Transpiration occurs at the leaf surface which is the loss of water due to the evaporation. This phenomenon works as trigger of water and mineral movement above to the xylem. Due to the evaporation of water at the leaf, negative pressure is created at the surface of leaf. Tension is produced which results in the pull of water from roots up to the xylem vessels.

You might be interested in

Find the relative molecular mass of hydrated iron (II) sulphate FeSO4.7H2O

Pepsi [2]

Answer:

the relative molecular mass of hydrated iron (II) sulfate FeSO4.7H2O is 278.02

Explanation:

5 0

3 years ago

What is the charge of Fe?

ki77a [65]

It depends on the context iorn is a transition metal so it can hold a charge from 1-8

7 0

3 years ago

An atom has three shells with 3 electrons in the last shell. This atom could be.....

Oksana_A [137]

Since the total amount of valence electrons is 3, it is in group 13 in the periodic table..therefore, it is specified as Boron.✅

4 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

4 A Small rock

vodomira [7]

3 millimeters is the rock

6 0

3 years ago