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2 years ago

Water droplets forming on a window on a cool day is an example of evaporation

Chemistry

2 answers:

Nataliya [291]2 years ago

7 0

Answer:

The statement is false.

Explanation:

Evaporation is a type of vaporization as the surface of a liquid changes into a gas. The event explained in the question refers to condensation, event that happens when a gas changes into a liquid state, such as the droplets forming on a window on a cool day.

Tomtit [17]2 years ago

5 0

Answer:

no its condensation

Explanation:

evaporation is when water.. well evaporated and condensation is when water condenses into liquid from a gas

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The two naturally occuring isotopes of antimony are 121Sb (57.21%) and 123Sb (42.79%), with isotopic masses of 120.904 and 122.9

emmasim [6.3K]

Answer:

The average atomic weight = 121.7598 amu

Explanation:

The average atomic weight of natural occurring antimony can be calculated as follows :

To calculate the average atomic mass the percentage abundance must be converted to decimal.

121 Sb has a percentage abundance of 57.21%, the decimal format will be

57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .

123 Sb has a percentage abundance of 42.79%, the decimal format will be

42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .

Next step is multiplying the fractional abundance to it masses

121 Sb = 0.5721 × 120.904 = 69.169178400

123 Sb = 0.4279 × 122.904 = 52.590621600

The final step is adding the value to get the average atomic weight.

69.169178400 + 52.590621600 = 121.7598 amu

5 0

3 years ago

How do you think the vertical temperature

WARRIOR [948]

thermal conduction and convection processes we explain the temperature profile: warmer at the bottom and colder at the top

Temperature is a measure of the internal energy of molecules.

When the pot is on the burner, the lower part of the pot acquires energy from the flame, this energy increases the temperature of the metal that is in contact with the liquid from the bottom, by a process of thermal conduction, while the liquid in the top is kept at the same temperature.

As time progresses, the hotter lower liquid, for which it has a lower with the metal bottom heats up by conduction. density, conscientiously rise and a transfer of heat is created by convection, which raises the temperature of the hot liquid and makes the liquid fall that is colder from the upper part than at the coming into contact

This cycle is repeated heating all the liquid. But the liquid with the liquid in the lower part always at a higher temperature than the liquid in the upper part, the process stops when the liquid in the lower part reaches the boiling point, than the passage from liquid to gas, during which the temperature remains constant.

With the thermal conduction and convection processes we explain the temperature profile: warmer at the bottom and colder at the top.

Learn more about temperature transfer here: brainly.com/question/24200572

4 0

3 years ago

How many grams of calcium chloride are needed to produce 1.50 g of potassium chloride? cacl2(aq) + k2co3(aq) → 2 kcl(aq) + caco3

Virty [35]

CaCl2(aq) + K2CO3(aq) → 2 KCl(aq) + CaCO3(aq)

1.12 g
2.23 g
0.896 g
4.47 g
1.12 g

Hope it helps :)

8 0

3 years ago

Read 2 more answers

How many grams of sodium phosphate monobasic would we add to a liter and how many grams of sodium phosphate dibasic would we add

Nostrana [21]

Answer :

The correct answer for Mass of Na₂HPO₄ = 4.457 g and mass of NaH₂PO₄ = 8.23 g

Given : pH = 6.86

Total concentration of Phosphate buffer = 0.1 M

Asked : Mass of Sodium phosphate monobasic (NaH₂PO₄) = ?

Mass of Sodium phosphate dibasic(Na₂HPO₄)= ?

Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :

(In phosphate buffer , Na+ ion from NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )

Step 1 : To find pka

H₂PO₄⁻ <=> HPO₄²⁻

The above reaction has pka = 7.2 ( from image shown )

Step 2 : Plug values in Hasselbalch- Henderson equation .

Hasselbalch -Henderson equation is to find pH for buffer solution which is as follows :

$pH = pka + log\frac{[A^-]}{[HA]}$

pH = 6.86 pKa = 7.2

$6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

Subtracting both side by 7.2

$6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

$-0.34 = log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

Removing log

$10^-^0^.^3^4 = \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

$\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457$ ---------------- equation (1)

Step 3 : To find molarity of H₂PO₄⁻ and HPO₄²⁻

Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M

Hence, [H₂PO₄⁻ ] + [ HPO₄²⁻ ] = 0.1 M

Assume [H₂PO₄⁻ ] = x

So , [x ] + [ HPO₄²⁻ ] = 0.1 M

[ HPO₄²⁻ ] = 0.1 - x

Step 4 : Plugging value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ]

[H₂PO₄⁻ ] = x

[ HPO₄²⁻ ] = 0.1 - x

Equation (1) = > $\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457$

Plug value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ] ( from step 3 ) into equation (1) as :

$\frac{[0.1 - x ]}{[ x]} = 0.457$

Cross multiplying

$0.1 - x = 0.457 x$

Adding x on both side

$0.1 -x + x = 0.457 x + x$

$0.1 = 1.457 x$

Dividing both side by 1.457

$\frac{0.1}{1.457} = \frac{1.457 x }{1.457}$

x = 0.0686 M

Hence , [H₂PO₄⁻ ] = x = 0.0686 M

[ HPO₄²⁻ ] = 0.1 - x

[ HPO₄²⁻ ] = 0.1 - 0.0686

[ HPO₄²⁻ ] = 0.0314 M

Step 5 : To find moles of H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .

Molarity is defined as mole of solute per 1 L volume of solution .

Molarity of NaH₂PO₄ = 0.0686 M or 0.0686 mole per 1 L

Molarity of Na₂HPO₄ = 0.0314 M or 0.0314 mole per 1 L

Since that volume of buffer solution is 1 L , so Molarity = mole

Hence Mole of NaH₂PO₄ = 0.0686 mol

Mole of Na₂HPO₄ = 0.0314 mol

Step 6 : To find mass of Na₂HPO₄ and NaH₂PO₄

Moles of Na₂HPO₄ and NaH₂PO₄ can be converted to their masses using molar mass as follows :

Molar mass of Na₂HPO₄ = $141.96 \frac{g}{mol}$

Molar mass of NaH₂PO₄ = $119.98 \frac{g}{mol}$

$Mass (g) = mole (mol)* molar mass(\frac{g}{mol})$

$Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}$

Mass of Na₂HPO₄ = 4.457 g

$Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}$

Mass of NaH₂PO₄ = 8.23 g

5 0

2 years ago

A student is asked to determine the chemical formula for the compound Barium

Sphinxa [80]

Answer:

Are barium and chlorine a metal and nonmetal?

Explanation:

I had the same question, and my class reviewed the test and this was the answer to this question.

5 0

2 years ago