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abruzzese [7]
3 years ago
14

H2O is a single atom of water

Chemistry
1 answer:
defon3 years ago
4 0

H_2O is defined as a single molecule of water.

<em>This is because it contains two hydrogen atoms and one oxygen atoms that are chemically bonded together.</em>

You might be interested in
What are air pollutants
KIM [24]

Answer:

things that pollute the air

Explanation:

nuclear energy, fossil fuels, gas from cars and other transportation, and these all contribute to greenhouse gases that are harmful to the environment

hope this helped!

5 0
2 years ago
Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O
Kipish [7]

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = \frac{15.0 g}{40 g/mol}=0.375 mol

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

Molarity=\frac{Moles}{Volume(L)}

n=0.250 M\times 0.150 L=0.0375 mol

NaOH+HNO_3\rightarrow NaNO_3+H_2O

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

\frac{1}{1}\times 0.0375 mol of NaOH

Moles of NaOH left unreacted  in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

1\times 0.3375 moles=0.3375 mol of hydroxide ion

The molarity of hydroxide ion in solution ;

=\frac{0.3375 mol}{0.150 L}=2.25 M

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

3 0
3 years ago
I NEED HELP ! ANYTHING WILL HELP!
raketka [301]
Trilobites are more common than Sea urchins
5 0
3 years ago
Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu
Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (<em>M)</em> is calculated from the following equation:

<em>M</em> = <em>n</em> ÷ <em>V</em>

where <em>n</em> is the number of moles of the solute in the solution, and <em>V</em> is the volume of the solution.

Accordingly, the number of moles of the solute is given by

<em>n</em> = <em>M</em> x <em>V</em>

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

<em>n </em>(1) = <em>M </em>(1) x <em>V </em>(1)

<em>n </em>(2) = <em>M </em>(2) x <em>V </em>(2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

<em>n</em> (1) = <em>n</em> (2)

and

<em>M</em> (1) x <em>V</em> (1)<em> </em>= <em>M</em> (2) x <em>V</em> (2)

For the stock solution, <em>M</em> (1) = 1.244 M, and <em>V</em> (1) is the volume needed. For the dilute solution, <em>M</em> (2) = 0,1281 M, and <em>V</em> (2) = 250.0 mL.

The volume of the stock solution needed, <em>V</em> (1), is calculated as follows:

<em>V</em> (1) = <em>M</em> (2) x <em>V</em> (2) ÷ <em>M</em> (1)

<em>V</em> (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

<em>V </em>(1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0
3 years ago
A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal
bixtya [17]

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V =  Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

8 0
3 years ago
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