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4 years ago

A. When 25.4 and 40.433 are added, the answer should be based on how significant figures or decimal

Chemistry

1 answer:

vladimir2022 [97]4 years ago

3 0

Answer:

65.8 and it is shown considering the significant figures of the shortest number.

Explanation:

Hello,

In this case, when we add 25.4 and 40.433 the result is 65.833, nevertheless, the result is shown based on the initial number having the fewest amount of significant figures that in this case is 25.4 that has three, therefore, the result must be 65.8 and it is shown considering the significant figures of the shortest.

Regards.

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White wash suspension, a saturated solution of ca(oh)2, has a ph of 12.37. what is ksp for ca(oh)2?

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Find the $pOH$ from $pH$

$pH + pOH =14\\pOH= 14-pH = 14-12.37 = 1.63$

The $KSP$ value of $ca(oh)^{2}$ is

$ca(oh)^{2} -- > ca^{2} + (aq) +2OH^{-} \\ksp = 4s^{3}$

The value finally becomes,

$ksp = 5.15 * x=10^{-5}$

<h3>What are the properties of Saturated solution ?</h3>

A saturated solution is one that contains all of the solute that can possibly dissolve in it. The most sodium chloride that may dissolve in $100 g$ of water at $20$ °C is $36.0 g$ . Past that point, adding more $NaCl$ won't cause it to dissolve because the solution is already saturated.

The solution reaches a point when it is saturated. This indicates that if you add more of the substance, it will stop dissolving and instead stay solid. The amount depends on how the solvent and solute interact molecularly.

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2 years ago

(LO 3N, 4G, 4O) Aluminum sulfate is also involved in dying fabrics. The gelatinous precipitate formed in the reaction with dilut

brilliants [131]

9.4 × 10⁻³ mg (0.73 mmoles) of Al(OH)₃ is formed

Explanation:

We have the following chemical reaction:

Al₂(SO₄)₃(aq) + 6 NaOH(aq) → 2 Al(OH)₃(s) + 3 Na₂SO₄(aq)

The precipitate mentioned by the problem is aluminium hydroxide Al(OH)₃.

Now to determine the number of moles of sodium hydroxide NaOH we use the following formula:

molar concentration = number of moles / volume

number of moles = molar concentration × volume

number of moles of NaOH = 0.088 M × 25 mL = 2.2 mmoles

number of moles of Al₂(SO₄)₃ = 5.6 × 10⁻³ moles = 5.6 mmoles (found in the problem text)

We see from the chemical reaction that 1 mole of Al₂(SO₄)₃ requires 6 moles of NaOH so 5.6 mmoles of Al₂(SO₄)₃ would require 6 times more NaOH which is 33.6 mmoles and we have only 2.2 mmoles. The limiting reactant will be NaOH.

Now we devise the following reasoning:

if 6 mmoles of NaOH produces 2 mmoles of Al(OH)₃

then 2.2 mmoles of NaOH produces X mmoles of Al(OH)₃

X = (2.2 × 2) / 6 = 0.73 mmoles of Al(OH)₃

mass of Al(OH)₃ = number of moles / molecular weight

mass of Al(OH)₃ = 0.73 / 78

mass of Al(OH)₃ = 9.4 × 10⁻³ mg

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3 years ago

What is the pressure of a gas in torr if its pressure is 305 kPa?

FromTheMoon [43]

Answer:

the pressure of the gas in torr is 2280 torr

Explanation:

The pressure of a gas in torr can be calculated by converting its pressure in kPa to atmospheres and then converting the atmospheric pressure to torr.

One atmosphere is equal to 101.325 kPa. Therefore, the pressure of the gas in atmospheres is 305 kPa / 101.325 kPa/atm = 3.00 atm.

One torr is equal to 1/760 of an atmosphere. Therefore, the pressure of the gas in torr is 3.00 atm * 760 torr/atm = 2280 torr.

Therefore, the pressure of the gas in torr is 2280 torr

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2 years ago

5K + KNO3 ------ 3K2O + N

masya89 [10]

Answer:

174,957.143 grams of potassium and 89,228.478 grams of potassium nitrate will be needed.

Explanation:

$5K +KNO_3\rightarrow 3K_2O + N$

Mass of nitrogen = 27 lbs = 12,247 g

1 lbs = 453.592 g

Moles of nitrogen = $\frac{12,247 g}{14 g/mol}=874.786 mol$

According to reaction, 1 mole of nitrogen is produced from 5 moles of potassium and 1 mole of potassium nitrate.

Then 874.786 mol of nitrogen will be obtained from :

$\frac{5}{1}\times 874.786 mol=4,373.928 mol$ of potassium.

Then 874.786 mol of nitrogen will be obtained from :

$\frac{1}{1}\times 874.786 mol=874.789 mol$ of potassium nitrate.

Mass of 4,373.928 moles of potassium:

$4,373.928 mol\times 40 g/mol=174,957.143 g$ of potassium

Mass of 874.789 moles of potassium nitrate:

$874.789 mol\times 102 g/mol=89,228.478 g$ of potassium nitrate

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3 years ago

Three positive charges lie on the x axis: q1 = 1 × 10−8 C at x1 = 1 cm, q2 = 2 × 10−8 C at x2 = 2 cm, and q3 = 3 × 10−8 C at x3

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Answer: Option (4) is the correct answer.

Explanation:

Relation between potential energy and charge is as follows.

U = $\frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]$

As it is given that $q_{1} = 1 \times 10^{-8} C$ , $q_{2} = 2 \times 10^{-8} C$ , and $q_{3} = 3 \times 10^{-8} C$ .

Distance between the charges = 1 cm = $1 \times 10^{-2} m$ (as 1 cm = 0.01 m)

Hence, putting these given values into the above formula as follows.

U = $\frac{1}{4 \pi \epsilon_{o}}[\frac{q_{1}q_{2}}{r_{12}} + \frac{q_{2}q_{3}}{r_{23}} + \frac{q_{3}q_{1}}{r_{31}}]$

= $9 \times 10^{9} [\frac{1 \times 10^{-8} \times 2 \times 10^{-8}}{10^{-2}} + \frac{2 \times 10^{-8} \times 3 \times 10^{-8}}{10^{-2}} + \frac{3 \times 10^{-8} \times 1 \times 10^{-8}}{10^{-2}}]$

= $9 \times 10^{9} [2 + 6 + 1.5]$

= $85.5 \times 10^{-5}$ J

= 0.00085 J

Thus, we can conclude that the potential energy of this arrangement, relative to the potential energy for infinite separation, is about 0.00085 J.

8 0

3 years ago