What type of reaction is this?—- Large Compound=Element A+Element B

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

During physical changes ,matter always retains its

svlad2 [7]

During physical changes, matter always retains its chemical properties.

7 0

3 years ago

Acid rain is a type of pollution caused mainly from

Aleksandr-060686 [28]

Answer:

im pretty sure its the second one

4 0

3 years ago

It is often necessary to prepare a dilute solution from a more concentrated solution rather than starting from a solid and just

valina [46]

Answer:

Take 10.0mL of your original solution and dilute it to 100.0mL and then, take 5.0mL of this solution and dilute it in another flask to 100.0mL

Explanation:

The initial solution of 0.3g of Red dye in 100mL has a concentration of:

0.3g Red dye × (1mol / 869g) = 3.4x10⁻⁴ mol Red Dye in 100mL = 0.100L:

3.4x10⁻⁴ mol / 0.1L = 3.4x10⁻³M

As you want a 1.7x10⁻⁵M solution, you need to dilute your solution:

3.4x10⁻³M / 1.7x10⁻⁵M = 200 times

Having this in mind, you must find the dilutions you must make to achieve the 200 of dilution factor (<em>Dilution factor is how many times you dilute a solution from a concetrated form).</em>

If you dilute the original solution taking 10.0mL to 100.0mL you dilute the solution: 100.0mL / 10.0mL = <em>10 times.</em>

Now, you need to dilute this solution another 20 times to achieve the 10*20 = 200 times of dilution.

If you take 5.0mL of the last solution and dilute it to 100.0mL you are diluting the solution 100.0mL / 5.0mL = <em>20 times.</em>

The dilution is:

3.4x10⁻³M × (10.0mL / 100.0mL) × (5.0mL / 100.0mL) = 1.7x10⁻⁵M

Thus, to obtain the 1.7x10⁻⁵M solutionyou need to:

<h3>Take 10.0mL of your original solution and dilute it to 100.0mL and then, take 5.0mL of this solution and dilute it in another flask to 100.0mL</h3>

8 0

3 years ago

The burning of fossil fuels results in which type of chemical reaction

Alika [10]

It results in a combustion reaction

5 0

3 years ago

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