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4 years ago

Examine the three isotopes of potassium (K) below.

Chemistry

1 answer:

liubo4ka [24]4 years ago

6 0

Answer: The correct answer is Potassium-39, Potassium-40, Potassium-41

Explanation:

Atomic number is defined as the number of protons present in an atom.

Mass number is defined as the sum of number of protons and number of neutrons present in an atom. It is represented as A.

A = Mass number = Number of neutrons + Number of protons.

For the given isotopes of potassium:

Atomic number of potassium is 19.

For potassium-41 isotope:

Mass number = 41

Number of neutrons = 41 - 19 = 22

For potassium-39 isotope:

Mass number = 39

Number of neutrons = 39 - 19 = 20

For potassium-40 isotope:

Mass number = 40

Number of neutrons = 40 - 19 = 21

The most number of neutrons are in potassium-41 isotope and the least number of neutrons are in potassium-39

Hence, the correct answer is Potassium-39, Potassium-40, Potassium-41

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If 5.0 grams of sucrose, C12H22O11, are dissolved in 10.0 grams of water, what will be the boiling point of the resulting soluti

GaryK [48]

Answer : The boiling point of the resulting solution is, $100.6^oC$

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

or,

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^oC$

$k_b$ = boiling point constant = $0.52^oC/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute (sucrose) = 5.0 g

$w_1$ = mass of solvent (water) = 10.0 g

$M_2$ = molar mass of solute (sucrose) = 342.3 g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^oC=1\times (0.52^oC/m)\times \frac{(5.0g)\times 1000}{342.3\times (10.0g)}$

$T_b=100.6^oC$

Therefore, the boiling point of the resulting solution is, $100.6^oC$

5 0

3 years ago

A chemist found a 60.00 gram sample containing copper and oxygen. Given the following percent composition, what is the mass of c

Strike441 [17]

The answer should 34.25g of Cu.

6 0

3 years ago

The yearly amounts of carbon emissions from cars in Belgium are normally distributed with a mean of 13.9 gigagrams per year and

zavuch27 [327]

Answer:

The correct option is;

c. 0.167

Explanation:

The parameters given are;

The mean, μ = 13.9 Gigagrams/year

The standard deviation, σ = 5.8 Gigagrams/year

The z-score formula is given as follows;

$z = \dfrac{x - \mu }{\sigma }$

Where:

x = Observed score = 11.5 Gigagrams/year

We have;

$z = \dfrac{11.5 - 13.9 }{5.8 } = \dfrac{-2.4 }{5.8 } = 0.4138$

From the z-score table relations/computation, the probability (p-value) = 0.6605

Where:

x = Observed score = 14 Gigagrams/year

We have;

$z = \dfrac{14- 13.9 }{5.8 } = \dfrac{0.1}{5.8 } = 0.01724$

From the z-score table relations/computation, the probability (p-value) = 0.4931

Therefore, the probability, $p_{ca}$ , that the amount of carbon emissions from cars in Belgium for a randomly selected year are between 15.5 Gigagrams/year and 14 Gigagrams/year = The area under the normal curve bounded by the p-values for the two amounts of carbon emission