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3 years ago

A scholar measures 20 mL of room temperature water (22ºC) and adds 80 mL of 75°C water to it, following the

Chemistry

1 answer:

Leni [432]3 years ago

4 0

Answer:

Data is not valid

Explanation:

When two liquids having different temperatures are mixed, regardless of the volumes, the final mix temperature will ALWAYS be between the initial temperature values.

1st Law Thermo => Law of Conservation of Energy => Energy can not be created nor destroyed, only changed in form. Mixing 22°C with 75°C will NOT result in a mix having a final temperature of 80°C.

∑ΔE = 0 => (mcΔT)₁ + (mcΔT)₂ = 0

[(20g)(1cal/g·°C)(Tₓ - 22°C)] + [(80g)(1cal/g·°C)(Tₓ - 75°C)] = 0

=> 20(Tₓ - 22) + 80(Tₓ - 75) = 0

=> 20Tₓ - 440 + 80Tₓ - 75 = 0

=> 100Tₓ = 440 + 75 = 515

=> Tₓ = (515/100)°C = 51.5°C final mix temperature

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How many moles are in 4.3 x 10 ^22 molecules of Na3PO4

Sever21 [200]

Answer:

0.071 moles of Na₃PO₄ .

Explanation:

Given data:

Number of molecules of Na₃PO₄ = 4.3× 10²² molecules

Number of moles = ?

Solution:

1 mole contain 6.022 × 10²³ molecules

4.3× 10²² molecules × 1 mol / 6.022 × 10²³ molecules

0.71× 10⁻¹ mol

0.071 mol

The number 6.022 × 10²³ is called Avogadro number.

"It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance"

3 0

3 years ago

Generally, systems move spontaneously in the direction of increasing entropy. TRUE FALSE

allochka39001 [22]

Answer:

true

Explanation:

5 0

2 years ago

What is the compound name of S3I9?

klasskru [66]

Trisulfur nonaiodide

Step-by-step explanation:
The name for a binary molecular compound has the form
Multiplying prefix+name of first element multiplying prefix+stem of second element element+ide (two words)
The multiplying prefixes for three and nine are tri and nona, respectively.
The stem of iodine is iod.
Put them together, and the name of S₃I₉ is
trisulfur nonaiodide.

4 0

2 years ago

Read 2 more answers

If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate

NARA [144]

Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

Na2S(aq) + Pb(NO3)2(aq) ----> 2NaNO3(aq) + PbS(s)

Ionically;

Pb^2+(aq) + S^2-(aq) -----> PbS(s)

Number of moles of sodium sulphide= concentration of sodium sulphide × volume of sodium sulphide

Number of moles of sodium sulphide= 0.780 × 1.5 = 1.17 moles

Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

Number of moles of lead II nitrate= 3.31× 1.00= 3.31 moles

Then we determine the limiting reactant. The limiting reactant yields the least amount of product.

Since 1 moles of sodium sulphide yields 1 mole of lead II sulphide

1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

4 0

3 years ago

What is the new pressure of 150 mL of a gas that is compressed to 50 mL when the original pressure was 3.0 ATM and the temperatu

Arisa [49]

Answer:

$\boxed {\boxed {\sf P_2= 9 \ atm}}$

Explanation:

Since the temperature is held constant, we only need to focus on the volume and pressure. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula is:

$P_1V_1=P_2V_2$

Originally, the gas had a volume of 150 milliliters and a pressure of 3.0 atmospheres. We can substitute these values into the left side of the equation.

$3.0 \ atm * 150 \ mL = P_2V_2$

The original gas was compressed to a volume of 50 milliliters, but we don't know the volume.

$3.0 \ atm *150 \ mL= P_2 * 50 \ mL$

Now, we need to solve for the new pressure (P₂). Multiply on the left side first.

$450 \ atm*mL= P_2 * 50 \ mL$

Since we are solving for the pressure, we need to isolate the variable. It is being multiplied by 50 mL. The inverse of multiplication is division. Divide both sides by 50 mL.

$\frac{450 \ atm*mL}{50 \ mL}= \frac{P_2 * 50 \ mL}{50 \ mL}$

$\frac{450 \ atm*mL}{50 \ mL}= P_2$

The units of milliliters will cancel.

$\frac{450 \ atm}{50 }= P_2$

$9 \ atm =P_2$

The new pressure is <u>9 atmospheres.</u>

8 0

3 years ago