How many tons is a gigaton of carbon?

How is the water cycle realated to weather patterns and climate? Make sure to put 2 evidences

Setler79 [48]

When there’s water in a department such as a lake, pound or/and ocean, the water tends to evaporated into the air and that’s how/when clouds are created. The same water from those locations are the same water that comes from the rain water, it’s a cycle that creates and takes water

7 0

3 years ago

Write the steps to separate the <br>Water from salt solution

zhuklara [117]

Answer:

you can use FILTRATION

STEP :-

1) put a hot water into breaker .

2) stir the solution until the sand dissolve

3) filtrate the solution

4) take the filtrate and evaporate under the sun

then you can get the water and salt too

Hope this helps :)

8 0

3 years ago

Read 2 more answers

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

Which of the following is NOT true about nuclear fusion?

inysia [295]

<u>Answer:</u>

I think it's (C)

The products are suitable for making nuclear weapons.

hope this helps!

6 0

3 years ago

How many protons and electrons are present in Au+?

balandron [24]

Answer:

Protons: 79

Electrons: 78

Explanation:

1. The number of protons is the atomic number (The atomic number for Au on the periodic table is 79)

2. Since the charge is +1 (positive) it means that there's one more proton than electrons. So, 79-1 = 78 electrons

7 0

3 years ago