Which of the following is the formula for Hypochlorous acid?

How does a difference in 1 pH unit change in terms of H+ concentration? Ex: How does a pH of 3 differ from the pH of 4? Which on

yuradex [85]

Answer:

A difference of 1 would mean there is either x10 more or less depending on where you are starting.

Explanation:

The pH of 3 differs from 4 because the acidity of 3 is 10x higher than that of 4. This is due to how the scientific notation of 3 in relation to pH is 10x^-3 while that of 4 is 10^-4 so in terms of the stronger acid that would be the one with the pH of 3.

6 0

3 years ago

Which one is a true statement about endothermic reactions? A. I need you from an outside source is continuously being added. B.

tekilochka [14]

Correct answer: "A. Energy from an outside source is continuously being added."
An endothermic reaction is a reaction that is characterised by the system absorbing energy from its surroundings. That energy is usually in heat form. For example, when mixing water<span> with potassium chloride, this reaction will absorb heat and the container will feel cold - endothermic reaction.</span>

7 0

3 years ago

Read 2 more answers

Please Help, will give 30 points! The following data was collected when a reaction was performed experimentally in the laborator

Sedaia [141]

Answer:

9 moles of NaNO3.

Explanation:

The balanced equation for the reaction is given below:

Al(NO3)3 + 3NaCl —> 3NaNO3 + AlCl3

Next, we shall determine the number of mole of Al(NO3)3 and NaCl that reacted and the number of mole of NaNO3 produced from the balanced equation.

From the balanced equation above,

1 mole of Al(NO3)3 reacted with 3 moles of NaCl to produce 3 moles of NaNO3.

Next, we shall determine the limiting reactant.

The limiting reactant can be obtained as follow:

From the balanced equation above,

1 mole of Al(NO3)3 reacted with 3 moles of NaCl.

Therefore, 4 moles of Al(NO3)3 will react with = (4 x 3)/1 = 12 moles of NaCl.

From the calculations made above, we can see that it will take a higher amount i.e 12 moles than what was given i.e 9 moles of NaCl to react completely with 4 moles of Al(NO3)3.

Therefore, NaCl is the limiting reactant and Al(NO3)3 is the excess reactant.

Finally, we shall determine the maximum amount of NaNO3 produced from the reaction.

In this case, the limiting reactant will be used as it will produce the maximum amount of NaNO3 since all of it is consumed by the reaction.

The limit reactant is NaCl and the maximum amount of NaNO3 produced can be obtained as follow:

From the balanced equation above,

3 moles of NaCl reacted to produce 3 moles of NaNO3.

Therefore, 9 moles of NaCl will also react to produce 9 moles of NaNO3.

From the calculations made above, the maximum amount of NaNO3 produced is 9 moles

6 0

3 years ago

A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature

Aliun [14]

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

$v_0=500$
$\alpha =1.12\times 10^{-4}$
$\Delta \epsilon = 25$

Now,

Increase in volume will be:

⇒ $\Delta V = \alpha\times v_0\times \Delta \epsilon$

By putting the given values, we get

$=1.12\times 10^{-4}\times 500\times 25$

$=1.12\times 10^{-4}\times 12500$

$=1.4 \ mL$

8 0

3 years ago

Sample A: 300 mL of 1M sodium chloride

yarga [219]

Answer:

sample B contains the larger density

Explanation:

Given;

volume of sample A, V = 300 mL = 0.3 L

Molarity of sample A, C = 1 M

volume of sample B, V = 145 mL = 0.145 L

Molarity of sample B, C = 1.5 M

molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol

Molarity is given as;

$C = \frac{moles \ of \ solute, \ mol}{liters \ of \ solvent} \\\\Moles \ of \ solute \ for \ sample \ A = 1 \times 0.3 = 0.3 \ mol\\\\Moles \ of \ solute \ for \ sample \ B = 1.5 \times 0.145 = 0.2175 \ mol$

The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g

The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g

The density of sample A $= \frac{mass}{volume} = \frac{17.55}{0.3} = 58.5 \ g/L$

The density of sample B $= \frac{mass}{volume} = \frac{12.72}{0.145} = 87.72 \ g/L$

Therefore, sample B contains the larger density

5 0

3 years ago