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3 years ago

Which is most likely to dissolve fastest in water? A. sugar cube B. powdered sugar

Chemistry

2 answers:

leva [86]3 years ago

7 0

Sugar cube is my answer

Hope this helps!
Let me know if i'm wrong.

lara31 [8.8K]3 years ago

7 0

The best answer is a) Sugar cube

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Describe one advantage and one disadvantage of replacing standard bulbs with led lighting

Leno4ka [110]

Advantages

less wasted energy

Reduces fossilisation

Disadvantages

- Expensive

- doesn't look nice ( apparently)

7 0

3 years ago

In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

<u>Answer:</u> The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

<u>Explanation:</u>

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

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

$M_1V_1=M_2V_2$ .......(1)

<u>Calculating for first dilution:</u>

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

<u>Calculating for second dilution:</u>

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

8 0

2 years ago

A 2.1−mL volume of seawater contains about 4.0 × 10−10 g of gold. The total volume of ocean water is about 1.5 × 1021 L. Calcula

Fiesta28 [93]

Answer:

Total worth of gold in the ocean = $5,840,000,000,000,000

Explanation:

As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.

Therefore, In 1 L of ocean water there will be,

(4.0 x 10^-10)/0.0021

= 1.9045 x 10^-7 g of gold per Liter of ocean water.

So in 1.5 x 10^-21 L of ocean water, there will be

(1.9045 x 10^-7) * (1.5 x 10^-21)

= 2.857 x 10^14 g of gold in the ocean.

1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is

20.44 * (2.857 x 10^14)

= $5,840,000,000,000,000

8 0

3 years ago

1) A mixture of anhydrous sodium carbonate and sodium hydrogencarbonate of mass 10.000 g was heated until it reached a constant

vodomira [7]

The masses of the components are obtained as;

Sodium hydrogen carbonate = 3.51 g
Sodium carbonate = 8.708 g

<h3>What is decomposition?</h3>

The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.

Now putting down the equation of the reaction, we have;

$2NaHCO_{3} (s) ----- > Na_{2} CO_{3} (s) + CO_{2} (g) + H_{2} O(g)$