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Mazyrski [523]
3 years ago
15

What keeps the water cycle going?

Chemistry
2 answers:
a_sh-v [17]3 years ago
6 0
The constant use of water flow throwout nature and the way that we use it now isint good and most likly we will be out of water by the year 2055
Mnenie [13.5K]3 years ago
3 0
The water cycle or hydrologic is a continuous cycle where water evaporates, travels into the air and becomes part of a cloud, falls down to earth as precipitation, and then evaporates again. ... Water keeps moving and changing from a solid to a liquid to a gas, over and over again.
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A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
Which coefficients balance the following equation?
Nookie1986 [14]
Answer: A
FeCl3 + 3 NaOH -> Fe(OH)3 + 3 NaCl
7 0
3 years ago
Compare<br> Ion and Radical<br> Atom and molecule<br> Organic and inorganic compounds
horrorfan [7]

Answer:

'An ion has a non-zero electric charge. A radical has an atom with unfilled electron shells and so is very reactive, but is electrically neutral.'

'Atoms are single neutral particles. Molecules are neutral particles made of two or more atoms bonded together.'

'The primary difference that lies between these organic compounds and inorganic compounds is that organic compounds always have a carbon atom while most of the inorganic compounds do not contain the carbon atom in them.'

6 0
2 years ago
Read 2 more answers
For the chemical reaction , I identify the reactant and the products .
svetlana [45]

The reactant is Mercury (II) Oxide while the products are Mercury and Oxygen separately.

This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.

Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.

Hope this helps!

7 0
4 years ago
how many molecules of sulfuric acid are in a spherical raindrop of diameter 6.0 mm if the acid rain has a concentration of 4.4 *
Vitek1552 [10]

Answer:

The number of moles =

Moles=4.97\times 10^{-8}

The number of molecules =

Molecules = 2.99\times 10^{16}

Explanation:

Volume of the sphere is given by :

V=\frac{4}{3}\pi r^{3}

here, r = radius of the sphere

radius=\frac{diameter}{2}

radius=\frac{6.0}{2}

Radius = 3 mm

r = 3 mm

1 mm = 0.01 dm (1 millimeter = 0.001 decimeter)

3 mm = 3 x 0.01 dm = 0.03 dm

r = 0.03 dm

<em>("volume must be in dm^3 , this is the reason radius is changed into dm"</em>

<em>"this is done because 1 dm^3 = 1 liter and concentration is always measured in liters")</em>

V=\frac{4}{3}\pi 0.03^{3}

V=\frac{4}{3}\pi 2.7\times 10^{-5}

V=1.13\times 10^{-4}dm^{3}

V=1.13\times 10^{-4}L   (1 L = 1 dm3)

Now, concentration "C"=

C=4.4\times 10^{-4}moles/liter  

The concentration is given by the formula :

C=\frac{moles}{Volume(L)}

This is also written as,

Moles = C\times Volume

Moles=1.13\times 10^{-4}\times 4.4\times 10^{-4}

Moles=4.97\times 10^{-8}moles

One mole of the substance contain "Na"(= Avogadro number of molecules)

So, "n"  mole of substance contain =( n x Na )

N_{a}=6.022\times 10^{23}

Molecules =

Molecule=4.97\times 10^{-8}\times 6.022\times 10^{23}

Molecules = 2.99\times 10^{16} molecules

7 0
3 years ago
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