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3 years ago

4. How would you calculate the poH of a base when given the concentration of [OH]-?

Chemistry

1 answer:

Novay_Z [31]3 years ago

3 0

Answer:

Use pOH = -log₁₀ [OH-]

Explanation:

pOH can be calculated from the concentration of hydroxide ions using the formular below:

pOH = -log₁₀ [OH-]

The pOH is the negative logarithm of the hydroxide ion concentration.

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Chemistry!! Please help!!

cluponka [151]

Answer:

4.993 ×10⁻¹¹ J

Explanation:

The nuclear binding energy is the energy equivalent to the mass defect.

The mass defect is the difference between the mass of a nucleus and the sum of the masses of its nucleons.

Calculate the mass defect

16 p = 16 × 1.007 28 u = 16.116 48 u

16 n = 16 × 1.008 67 u = 16.138 72 u

Total mass of nucleons = 32.255 20 u

- Mass of S-32 = 31.972 070 u

Mass defect = 0.283 13 u

Convert the unified atomic mass units to kilograms.

Mass defect

$= \text{0.283 13 u} \times \frac{ \text{1.66 06} \times 10^{-27} \text{ kg}}{\text{1 u}}$

$= 4.700 \times 10^{-28} \text{ kg}$

Use Einstein’s equation to convert the mass defect into energy

$E = mc^{2}$

$E = 4.7000 \times 10^{-28} \text{ kg} \times (2.998 \times 10^{8} \text{ m}\cdot\text{s}^{-1})^{2} = 4.224 \times 10^{-11} \text{ J}$

4 0

4 years ago

I NEED HELP PLEASE !!!!!

kolbaska11 [484]

Answer: jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

Explanation:

4 0

3 years ago

A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?

Viktor [21]

The answer for the following problem is mentioned below.

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.

Explanation:

Given:

mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 125.3 grams

We know;

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 + 3(95)

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 +285 = 405 grams

We also know;

No of molecules at STP conditions( $N_{A}$ ) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

$\frac{m}{M} =\frac{N} }{}$ N÷ $N_{A}$

$\frac{405}{125.3}$ = $\frac{N}{6.023*10^23}$

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules

3 0

3 years ago

What volume of a 2.0M NaOH(aq) is needed to completely neutralize 24 milliliters of 0.5M HCl(aq)?

borishaifa [10]

Answer:

$V_{base}=6.0mL$

Explanation:

Hello there!

In this case, by considering that the reaction between sodium hydroxide and hydrochloric acid is in a 1:1 mole ratio of these two reactants, we are able to use the following equation relating the concentration and volume of each one:

$M_{acid}V_{acid}=M_{base}V_{base}$