Which term names the main class of the chemical that makes up bread.

BY ANSWERING THIS QUESTION UR PUTTING IT ON UR MOM's LIFE THAT U WON'T STEAL MY POINTS.

Yakvenalex [24]

Answer:

$T_2=-125.58\°C$

Explanation:

Hello!

In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:

$\frac{T_1}{P_1} =\frac{T_2}{P_2}$

Whereas solving for the final temperature T2, we get:

$T_2=\frac{T_1P_2}{P_1}$

Thus, we plug in the given data (temperature in Kelvins) to obtain:

$T_2=\frac{(22+273.15)K*1.75atm}{3.50atm} \\\\T_2=147.58K-273.15\\\\T_2=-125.58\°C$

Best regards!

3 0

2 years ago

What is the volume of 0.80 grams of O2 gas at STP? (5 points) Group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l

Anarel [89]

Answer:

0.56 liters

Explanation:

First we <u>convert 0.80 grams of O₂ into moles</u>, using its molar mass:

0.80 g ÷ 32 g/mol = 0.025 mol

At STP, 1 mol of any given mass occupies 22.4 L. With that information in mind we <u>calculate the volume that 0.025 moles of O₂ gas would occupy</u>:

0.025 mol * 22.4 L/mol = 0.56 L

Thus the answer is 0.56 liters.

3 0

2 years ago

The statement that percent yield can never be greater than theoretical yield is another example of the ________.

Gnom [1K]

We can rephrase the statement with a little more specificity in order to understand the answer here.

The mass of the products can never be more than the The mass that is expected.

3 0

3 years ago

Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simulta

scoray [572]

Answer:

70.77 g/mol is the molar mass of the unknown gas.

Explanation:

Effusion is defined as rate of change of volume with respect to time.

Rate of Effusion= $\frac{Volume}{Time}$

Effusion rate of oxygen gas after time t = $E=\frac{4.64 mL}{t}$

Molar mass of oxygen gas = M = 32 g/mol

Effusion rate of unknown gas after time t = $E'=\frac{3.12 mL}{t}$

Molar mass of unknown gas = M'

The rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

$\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$