How many protons neutrons and electrons are there in a neutral atom of 43k (potassium-43)?

684 kcal. One mole of glucose weighs roughly 180g. Given that 1g of glucose releases 3.8 kcal, we calculate for 1 mole of glucose: 180g -> 180g * 3.8 kcal/g = 684 kcal.

Answer:

Explanation:

AgNO3 + NaCl --> AgCl + NaNO3

Moles

of AgNO3

= molarity * volume

= 1 * 0.01

= 0.01 mol

for NaCl

= 0.01 * 1

= 0.01 mol.

According to stoichiometry, one mole of silver nitrate corresponds to one mole of NaCl reacted. Hence,

Moles of AgCl generated = 0.01 × 1

= 0.01 mol AgCl produced.

Heat gained by the solution as precipitation occurs:

Solution mass = density × volume

= 1 × 20

= 20 g.

Using q = m * Cp * (T2 - T1)

= 20 * 4.18 * (32.6 - 25.0)

= 635 J

The absorbed heat of 635 J indicates the reaction released -635 J

Thus, Delta H = -635 J/0.01 mol

= -63500 J/mol

= -63.5 kJ/mol.

Answer: A= 0.5, B = 0.25, C = 0.125, D = 0.015625 and E = 0.00390625

Explanation:

The half-life of a substance refers to the duration it takes for that substance to decrease to half of its original quantity.

Here, n signifies the count of half-lives.

The remaining quantity of the substance after n half-lives can be determined using the formula provided,

When n = 1,

Fraction of substance remaining = 0.5¹ = 0.5.

This indicates that after the first half-life, the remaining quantity is half of the original amount.

Thus, we find A = 0.5

<pfor n="2," we="" calculate=""><pso after="" two="" half-lives="" the="" remaining="" quantity="" is="" of="" original=""><phence>B = 0.25 <pfor n="3," we="" have=""><pso after="" three="" half-lives="" the="" remaining="" amount="" is="" of="" original.=""><ptherefore we="" find="">C = 0.125<pfor n="6," we="" see="">

Thus, D = 0.015625

<pfor n="8," we="" calculate=""><pconsequently>E = 0.00390625

The calculated values for A, B, C, D, and E are 0.5, 0.25, 0.125, 0.015625, and 0.00390625 respectively.

</pconsequently></pfor></pfor></ptherefore></pso></pfor></phence></pso></pfor>

Answer:

7.3

Explanation:

Using the Henderson Hasselbalch equation, one can determine the pH or pOH of a solution via its pKa. Remember, , and pKa = -logKa, where Ka denotes the acid's equilibrium constant.

The Henderson Hasselbalch formula:

In this context, acid X possesses two ionic forms: the carboxyl group and an alternative form. Initially, we have 0.1 mol/L of acid in 100 mL, which gives:

n1 = (0.1 mol/L)×(0.1 L) = 0.01 mol

Upon dissociation, it yields 0.005 mol of the carboxyl form and 0.005 mol of the other form with stoichiometry assumed constant.

Introducing NaOH at a concentration of 0.1 mol/L and 75 mL, the moles of become:

n2 = (0.1 mol/L)×(0.075 L) = 0.0075 mol

Thus, 0.0075 mol of reacts with 0.005 mol of the carboxyl form, leading to 0.0025 mol of , which in turn reacts with 0.005 mol of the alternating group, leaving 0.0025 mol of the latter.

The new solution’s volume is 175 mL, but the concentrations of both forms remain unchanged in volume, so we can utilize the moles in the equation.

<pNote, the moles of the acid form remain 0.01 mol as it doesn’t undergo reaction!

Thus, we arrive at:

6.72 = pKa - log 4

pKa - log 4 = 6.72

pKa = 6.72 + log 4

pKa = 6.72 + 0.6

pKa = 7.3

Science is ever-evolving with continuous discoveries; thus, the statement "it is stable and does not require modification due to new findings" is incorrect.