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Viefleur [7K]
2 years ago
14

Anyone (cool, funny, middle school) wanna start a conversation?

Chemistry
2 answers:
pychu [463]2 years ago
5 0

Answer:

sure

Explanation:

what's up

nadya68 [22]2 years ago
3 0

Answer:

Im down, Im in 7th

Explanation:

sure hi lol

You might be interested in
Mandy made the following table describing the conditions required to form sleet and snow, but left something out. Which word sho
MakcuM [25]

I am not the best at this, but im pretty sure its below cloud

7 0
2 years ago
A gas has a volume of 4.25 m3 at a temperature of 95.0°C and a pressure of 1.05 atm. What temperature will the gas have at a pre
Goryan [66]

Answer:

\boxed {\boxed {\sf 82.7 \textdegree C}}

Explanation:

We are asked to find the temperature of a gas given a change in pressure and volume. We will use the Combined Gas Law, which combines 3 gas laws: Boyle's, Charles's, and Gay-Lussac's.

\frac {P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Initially, the gas has a pressure of 1.05 atmospheres, a volume of 4.25 cubic meters, and a temperature of 95.0 degrees Celsius.

\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{P_2V_2}{T_2}

Then, the pressure increases to 1.58 atmospheres and the volume decreases to 2.46 cubic meters.

\frac {1.05 \ atm * 4.25 \ m^3}{95.0 \textdegree C}= \frac{1.58  \ atm *2.46 \ m^3}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. Cross multiply. Multiply the first numerator by the second denominator, then multiply the first denominator by the second numerator.

(1.05 \ atm * 4.25 \ m^3) * T_2 = (95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)

Now the variable is being multiplied by (1.05 atm * 4.25 m³). The inverse operation of multiplication is division, so we divide both sides by this value.

\frac {(1.05 \ atm * 4.25 \ m^3) * T_2}{(1.05 \ atm * 4.25 \ m^3)} = \frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}

T_2=\frac{(95.0 \textdegree C)*(1.58 \ atm * 2.46 \ m^3)}{(1.05 \ atm * 4.25 \ m^3)}

The units of atmospheres and cubic meters cancel.

T_2=\frac{(95.0 \textdegree C)*(1.58* 2.46 )}{(1.05 * 4.25 )}

Solve inside the parentheses.

T_2= \frac{(95.0 \textdegree C)*3.8868}{4.4625}

T_2= \frac{369.246}{4.4625} \textdegree C}

T_2 = 82.74420168 \textdegree C

The original values of volume, temperature, and pressure all have 3 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 4 in the hundredth place to the right tells us to leave the 7 in the tenths place.

T_2 \approx 82.7 \textdegree C

The temperature is approximately <u>82.7 degrees Celsius.</u>

3 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0 mL of 1 M H2SO4. A 25.00 mL aliquot is anal
Olenka [21]

Answer:

The weight percent in the sample is 17,16%

Explanation:

The dissolution of the Ce(IV) salt provides free Ce⁴⁺ that reacts, thus:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃²⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03247M Na₂S₂O₃ = 4,228x10⁻⁴ moles of S₂O₃²⁻.

As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,228x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^{2-}} = <em>2,114x10⁻⁴ moles of I₃⁻</em>

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,114x10⁻⁴ moles of I₃⁻× \frac{2molCe^{4+}}{1molI_{3}^-} =  <em>4,228x10⁻⁴ moles of Ce(IV)</em>.

These moles are:

4,228x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = 0,05924 g of Ce(IV)

As was taken an aliquot of 25,00mL from the solution of 250,0mL:

0,05924 g of Ce(IV)×\frac{250,0mL}{25,00mL} =0,5924g of Ce(IV) in the sample

As the sample has 3,452g, the weight percent is:

0,5924g of Ce(IV) / 3,452g × 100 = <em>17,16 wt%</em>

I hope it helps!

3 0
3 years ago
Humanities !
ella [17]

Which ancient Greek philosopher was the focus of study for the Italian humanists?

ANS : Aristotle

5 0
2 years ago
10. A small gold nugget has volume of 0.87 cm3. What is its mass if the density of gold is 19.3 g/cm3?
bulgar [2K]

Answer:

16.791 grams

Explanation:

The density formula is:

d=\frac{m}{v}

Rearrange the formula for m, the mass. Multiply both sides of the equation by v.

d*v=\frac{m}{v}*v

d*v=m

The mass of the gold nugget can be found by multiplying the density and volume. The density is 19.3 grams per cubic centimeter and the volume is 0.87 cubic centimeters.

d= 19.3 g/cm^3\\v-0.87 cm^3

Substitute the values into the formula.

m=d*v

m= 19.3 g/cm^3*0.87 cm^3

Multiply. Note that the cubic centimeters, or cm³ will cancel each other out.

m=19.3 g*0.87

m=16.791 g

The mass of the gold nugget is 16.791 grams.

8 0
3 years ago
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