(BRAINLIEST)<br> Which of these statements best describes an Alkali?

Artyom0805 [142]

Answer: A. To change the potential energy of the reactants.

Explanation:
Catalysts make such a breaking and rebuilding happen more efficiently. They do this by lowering the activation energy for the chemical reaction. Activation energy is the amount of energy needed to allow the chemical reaction to occur. The catalyst just changes the path to the new chemical partnership.

5 0

3 years ago

Read 2 more answers

Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.

maria [59]

The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%

5 0

2 years ago

Gamma rays are negatively charged.

asambeis [7]

Answer:

True

Explanation:

8 0

3 years ago

What is the [OH-] of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 m

velikii [3]

Answer: The $[OH^-]$ of a solution is $10^{-12}$ M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in ml

moles of $HCl$ = $\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0912g}{36.5g/mol}=0.0025mol$

Now put all the given values in the formula of molality, we get

$Molarity=\frac{0.0025\times 1000}{250}=0.01$

pH or pOH is the measure of acidity or alkalinity of a solution.

$HCl\rightarrow H^++Cl^{-}$

According to stoichiometry,

1 mole of $HCl$ gives 1 mole of $H^+$

Thus $0.01$ moles of $HCl$ gives = $\frac{1}{1}\times 0.01=0.01$ moles of $H^+$

Putting in the values:

$[H^+][OH^-]=10^{-14}$

$[0.01][OH^-]=10^{-14}$

$[OH^-]=10^{-12}$

Thus the $[OH^-]$ of a solution prepared by dissolving 0.0912 g of hydrogen chloride in sufficient pure water to prepare 250.0 ml of solution is $10^{-12}$ M

8 0

3 years ago

A student dissolves 6.2g of aniline c6h5nh2 in 350.ml of a solvent with a density of 1.04/gml . the student notices that the vol

Troyanec [42]

Answer : The correct answer for molarity = 0.19 $\frac{mol}{L}$ and Molality = 0.18 $\frac{mol}{Kg}$ .

Given : Mass of aniline = 6.2 g

Volume of solvent = 350 mL Density of solvent = 1.04 g/mL

1) Molarity : It is defined as number of moles of solute present in Litre of solution . It is expressed as :

$Molarity(M) = \frac{moles of solute(mole)}{volume of solution (L)}$ \

Molairty can be found in following steps :

Step 1 : To calculate mole :

Mole of solute can be calculate using mole formula :

$Mole of solute = \frac{given mass of solute (g)}{molar mass of solute\frac{g}{1 mol}}$

Molar mass of aniline (C₆H₅NH₂) = 93.13 $\frac{g}{1 mol}$

Mass of aniline = 6.2 g

Plugging values in mole formula :

$Mole = \frac{6.2 g}{93.13 \frac{g}{1 mol}}$

Mole = 0.0665 mol

Step 2 : To find volume of solution

Volume of solution = volume of solute + volume of solution

Since addition of aniline does not change final volume , so volume of solvent = volume of solution. Since volume is given in mL , so it need to be converted to L .

1 L = 1000mL

$Volume of solution = \frac{350 mL}{1000mL} * 1 L$