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Ilya [14]
4 years ago
14

Under which conditions will sugar most likely dissolve fastest in a cup of water

Chemistry
2 answers:
Sophie [7]4 years ago
5 0
Heated, disruption (mixing the solution), large surface area ( the sugar is ground up, not in a cube)
qwelly [4]4 years ago
4 0
It will dissolve fastest in a cup of hot water. 
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Why do we need continual input of energy in the body?
OverLord2011 [107]

Answer:

To move? I think that's right

7 0
3 years ago
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2' A mixture containing 2.75 gof ammonium chloride (NH4cl) in 5.0 g of water was heated to dissolve the solid and then allowed t
Trava [24]

Answer:

55g NH₄Cl / 100g Water

Explanation:

Solubility of a substance define the amount of solute per solvent in a saturated solution. The solution can dissolve additional solute if heated.

In the problem, as the first crystal appears at 61°C the solubility in this temperature is the concentration of the solution, that is:

2,75g NH₄Cl / 5,0g water ₓ 100 = <em>55g NH₄Cl / 100g Water</em>

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I hope it helps!

6 0
3 years ago
The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: CO(g) + H2O(g)
taurus [48]

<u>Answer:</u> The concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

<u>Explanation:</u>

For the given chemical reaction:

CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

The expression of K_c for above reaction follows:

K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}         ........(1)

We are given:

[CO]_{eq}=[H_2O]_{eq}=[H_2]_{eq}=0.10M

[CO_2]_{eq}=0.40M

Putting values in above equation, we get:

K_c=\frac{0.40\times 0.10}{010\times 0.10}\\\\K_c=4

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of hydrogen gas = 0.30 mol

Volume of solution = 2.0 L

Putting values in above equation, we get:

\text{Molarity of }H_2=\frac{0.30mol}{2L}=0.15M

When hydrogen gas is added, the concentration of product gets increased. But, by Le-Chatelier's principle, the equilibrium will shift in the direction where concentration of product must decrease, which is in the backward direction.

Concentration of hydrogen gas when equilibrium is re-established = 0.1 + 0.15 = 0.25 M

Now, the equilibrium is shifting to the reactant side. The equation follows:

                      CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial:              0.1      0.1                 0.4       0.1

At eqllm:       0.1+x   0.1+x           0.4-x      0.25-x

Putting values in expression 1, we get:

4=\frac{(0.25-x)(0.4-x)}{(0.1+x)(0.1+x)}\\\\3x^2+1.45x-0.06=0\\\\x=0.038,-0.522

Neglecting the negative value of 'x'

Calculating the concentrations of the species:

Concentration of carbon dioxide = (0.4 - x) = (0.4 - 0.038) = 0.362 M

Concentration of hydrogen gas = (0.25 - x) = (0.25 - 0.038) = 0.212 M

Concentration of carbon monoxide = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Concentration of water = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Hence, the concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

8 0
4 years ago
Two carbon atoms cannot be linked to each other by more than three covalent bonds why​
Leto [7]

Answer:since,there is a single bond between the two carbon atoms and both share their one atom therefore for completing its shell it need to combine with three atoms of carbon or other element . therefore it cannot be linked to more than 3 covalent bonds since its shell will be completed to become stable .

Explanation:

6 0
3 years ago
A student is given an antacid tablet that weighed 5.6832 g. The tablet was crushed and 4.3628 g of he antacid was added to 200 m
SSSSS [86.1K]

Answer:

a) 8.33 ml of the original stomach acid is neutralized

b) 191.67 ml of the stomach acid was neutralized

c)  249.68 ml acid would be neutralized by the original tablet

Explanation:

a) how much of the stomach acid had been neutralized in the 25 mL sample wich was titrated?

25.5 ml of a NaOH solution is equivalent to 25.00 ml of the original stomach acid

8.5 ml NaOH * (25.00 ml original stomach acid / 25.5 ml NaOH) = 8.33 ml original stomach acid

b) how much stomach acid was neutralized y the 4.3628 g tablet?

It takes 8.5 ml NaOH to neutralize 8.33 ml original acid (this is the answer for question 1)

This means the antacid neutralized = 200 ml - 8.33 ml = 191.67 ml

c) how much stomach acid would have been neutralized by the original 5.6832 g tablet

4.3628 g antacid is equivalent to 191.67 ml acid  ( this is the answer for question 2)

5.6832g antacid * (191.67 ml acid / 4.3628 g antacid) = 249.68 ml acid

3 0
3 years ago
Read 2 more answers
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