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2 years ago

Note the yellow lines which show attractions between the molecules. Why do you think these attractions occur?

Chemistry

1 answer:

ycow [4]2 years ago

8 0

A hydrogen bond (H-bond) is an example of intermolecular force (IMF).

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Nuclear energy could come from

lianna [129]

Nuclear energy comes from splitting of Barium atom to form Krypton atom

5 0

3 years ago

The net ionic equation for formation of an aqueous solution of nii2 accompanied by evolution of co2 gas via mixing solid nico3 a

Ann [662]

Answer:

NiCO3 (s) + 2H+ (aq) → H2O (l) + CO2 (g) + Ni2+ (aq)

Explanation:

To write the complete ionic equation:

1. Start with a balanced molecular equation.

2. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions

3. indicate the correct formula and charge of each ion

4. indicate the correct number of each ion

5. write (aq) after each ion

6. Bring down all compounds with (s), (l), or (g) unchanged.

8 0

3 years ago

Sodium carbonate (NaCO3) is sometimes used as a water-softening agent. Suppose that a worker prepares a 0.730 M solution of NaCO

elena55 [62]

Answer : The mass of solute in solution is $1.09\times 10^2g$ .

Solution : Given,

Molarity = 0.730 M

Volume of solution = 1.421 L

Molar mass of sodium carbonate = 105.98 g/mole

Formula used for Molarity :

$Molarity=\frac{w}{M\times V}$

where,

w = mass of solute

M = Molar mass of solute

V = volume of solution in liter

Sodium carbonate is solute and water is solvent.

Now put the given values in above formula, we get the mass of solute in solution.

$0.730mole/L=\frac{w}{(105.98g/mole)\times (1.412L)}$

By rearranging the terms, we get

$m=109.247g=1.09\times 10^2g$

Therefore, the mass of solute in solution is $1.09\times 10^2g$ .

4 0

2 years ago

What is solution in chemistry

nika2105 [10]

a solution in chemistry is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.

4 0

3 years ago

A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pres

tatiyna

Answer:

The partial pressure of argon in the flask = 71.326 K pa

Explanation:

Volume off the flask = 0.001 $m^{3}$

Mass of the gas = 1.15 gm = 0.00115 kg

Temperature = 25 ° c = 298 K

Gas constant for Argon R = 208.13 $\frac{J}{kg k}$

From ideal gas equation P V = m RT

⇒ P = $\frac{m R T}{V}$

Put all the values in above formula we get

⇒ P = $\frac{0.00115}{0.001}$ × 208.13 × 298

⇒ P = 71.326 K pa

Therefore, the partial pressure of argon in the flask = 71.326 K pa

4 0

3 years ago