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denis-greek [22]
3 years ago
10

If 180 grams of potassium iodide is dissolved in 100 cm3 of water at 30oC, a(n) _______________ solution is formed.

Chemistry
2 answers:
olga nikolaevna [1]3 years ago
6 0

Super saturated solution is formed.

<u>Explanation:</u>

Solubility is the property of any substance's capacity, that is the solute of the substance is dissolved in the given solvent to form the solution. We have three different types of solution, unsaturated, saturated and supersaturated solution.

  • Unsaturated solution is a solution with lesser amount of solute than its solubility at equilibrium.
  • Saturated solution is a solution with the maximum solute dissolved in the solvent.
  • Super saturated solution is a solution with more solute than it is required.

The solubility of KI at 30°C is 153 g / 100 ml. Here 180 g of KI in 100 ml of water at 30°C is given, which has more solute than required, so it is super saturated solution.

hjlf3 years ago
4 0

Answer:

What they are trying to say is that its answer choice "D".

Explanation:

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The half life of tritium is 12.3 years. How much sample will remain if you start with 425g of tritium that is found after 4 half
natta225 [31]

Answer:

26.56g

Explanation:

The half life of an isotope decay is the time transcurred when the initial amount of the isotope decays to the half of its amount.

For the problem, when 1 half life transcur, the sample has a mass of:

425g / 2 = 212.5g

When 2 half lives transcur:

212.5g / 2 = 106.25g

3 half lives:

106.25/2 = 53.13g

And the amount that is found after 4 half lives is:

53.13/2 =

<h3>26.56g</h3>
3 0
3 years ago
What is the molarity of a sodium carbonate solution containing 32.52 g Na2CO3 dissolved in enough water to make 822 ml of soluti
svet-max [94.6K]

The molarity of a Sodium carbonate solution : 0.373 M

<h3>Further explanation</h3>

Given

32.52 g Na₂CO₃

822 ml of solution = 0.822 L

Required

The molarity

Solution

Molarity shows the number of moles of solute in every 1 liter of solution.

\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}

mol of solute = mol of Na₂CO₃ :

= mass : MW Na₂CO₃

= 32.52 g : 106 g/mol

= 0.307

Molarity :

= n : V

= 0.307 mol : 0.822 L

= 0.373 M

8 0
3 years ago
Consider the following reaction: 2 H2(g) + 2 NO(g) → 2 H2O(g) + N2(g) If the concentration of NO changed from 0.100 M to 0.025 M
Wewaii [24]

Answer: The average rate of the reaction during this time interval is, 0.005 M/s

Explanation:

The given chemical reaction is:

2H_2(g)+2NO(g)\rightarrow 2H_2O(g)+N_2(g)

The expression will be:

\text{Average rate of reaction}=\frac{C_2-C_1}{t_2-t_1}

where,

C_2 = final concentration of NO = 0.025 M

C_1 = initial concentration of NO = 0.100 M

t_2 = final time = 15 minutes

t_1 = initial time = 0 minutes

Putting values in above equation, we get:

\text{Average rate of reaction}=\frac{0.025-0.100}{15-0}

\text{Average rate of reaction}=0.005M/s

Hence, the average rate of the reaction during this time interval is, 0.005 M/s

6 0
3 years ago
Consider the substances hydrogen (H2), fluorine (F2), and hydrogen fluoride (HF). Based on their molecular structures, how does
katovenus [111]

Answer:

The boiling point of HF is <u><em>higher than</em></u> the boiling point of H2, and it is <u><em>higher than</em></u> the boiling point of F2.

Explanation:

In HF, inter- molecule forces will be present between the hydrogen and fluorine atoms. There will be hydrogen bonding present among the hydrogen and fluorine atoms. Hydrogen bonds are strong bonds and hence the boiling point for HF would be high as much energy will be required to break these bonds.

H2 and F2 will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.

3 0
3 years ago
Coal like graphite is composed of carbon. Unlike graphite, coal is formed by decayed, once living matter. Is it a mineral?
Simora [160]
Coal is not a mineral. It's not a mineral because minerals are inorganic. Coal is organic because it comes from decayed living matter. 
3 0
3 years ago
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