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amid [387]
3 years ago
6

Calculate the volume of naoh solution that contains 0.72 g of a 8.0 % (m/v) naoh solution.

Chemistry
1 answer:
Hitman42 [59]3 years ago
3 0
A solution contains 12 g of glucose in 240 mL of solution. What is the mass/volume % of the solution? 5.0% (m/v). Calculate the grams of KCl in 225 g of an 8.00 % (m/m) KCl solution
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What occurs in order to break the bond in a Cl, molecule?
Ann [662]

Answer:

Explanation:

The attachment or (Bond) that resolves around the atoms of CI2 which is a covalent bond, yet energy is needed to destroy the bond and when new bonds are produced, energy is unleashed while the bond's development makes the molecule equilibrated. In short terms to basically break this bond energy is required and another terms energy is occupied and is the conclusion to your question (I tried my best) Energy is absorbed which is the result of what occurs.

8 0
3 years ago
How many moles are present in 356.4 g of NiBr3
lozanna [386]

1.194 mol

(remember to use sig figs!)

8 0
3 years ago
Consider 4.60 L of a gas at 365 mmHg and 20 C . If the container is compressed to 2.60 L and the temperature is increased to 36
BaLLatris [955]

Explanation:

The given data is as follows.

       V_{1} = 4.60 L,        P_{1} = 365 mm Hg

       V_{2} = 2.60 L,       P_{2} = ?

       T_{1} = (20 + 273) K = 293 K,      T_{2} = (36 + 273) K = 309 K

Since, number of moles of gas are equal so, according to ideal gas equation:

          \frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

          \frac{365 mm Hg \times 4.60 L}{293 K} = \frac{P_{2} \times 2.60 L}{309 K}

          P_{2} = 681.03 mm Hg

Thus, we can conclude that new pressure P_{2} is 681.03 mm Hg.

6 0
3 years ago
The mercury content of a stream was believed to be above the minimum considered safe—1 part per billion (ppb) by weight. an anal
shutvik [7]

Answer:

Amount of mercury is 1.0*10⁻⁵ g

Explanation:

<u>Given:</u>

Mercury content of stream = 0.68 ppb

volume of water = 15.0 L

Density of water = 0.998 g/L

<u>To determine:</u>

Amount of mercury in 15.0 L of water

<u>Calculation:</u>

1 ppb = \frac{1\mu g(solute)}{1L(solvent)}

where 1 μg (micro gram) = 10⁻⁶ g

0.68 ppm implies that there is 0.68 *10⁻⁶ g mercury per Liter of water

Therefore, the amount of mercury in 15.0 L water would be:

=\frac{0.68*10^{-6}g\ Hg* 15.0\ L\ water}{1\ L\ water} =1.02*10^{-5}g

4 0
3 years ago
Will give 50 points, If a decrease in temperature accompanies a reaction, what occurred?
Ede4ka [16]

Answer:

A.

Explanation:

3 0
3 years ago
Read 2 more answers
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