HELP HELP HELP HOW MANY GRAMS OF SUGAR CAN DISSOLVE IN 100 GRAMS OF WATER HEATED TO 40 CELCIOUS 60 POINTS OFFERED

Cloud [144]

Answer:

The definition of air pressure is the force exerted onto a surface by the weight of the air. An example of air pressure is the average sea-level air pressure of 101.325

Explanation:

7 0

2 years ago

Use the definition of molarity to calculate the concentration of 12.34 g of CaSO4 completely dissolved in water, with a solution

Ede4ka [16]

Answer:

[CaSO₄] = 36.26×10⁻² mol/L

Explanation:

Molarity (M) → mol/L → moles of solute in 1L of solution

Let's convert the volume from mL to L

250 mL . 1L/1000 mL = 0.250L

We need to determine the moles of solute. (mass / molar mass)

12.34 g / 136.13 g/mol = 0.0906 mol

M → 0.0906 mol / 0.250L = 36.26×10⁻² mol/L

8 0

2 years ago

HELP! ASAP!!! WILL MARK BRAINLIEST!!

san4es73 [151]

Answer:

Rb

Alkali Metals are Group 1 so

Rb it isnt Hydrogen because it is a gas

Explanation:

8 0

2 years ago

Read 2 more answers

A gas takes up a volume of 10 liters, has a pressure of 7.5 atm, and a

Y_Kistochka [10]

Answer:

I’m trying to do something similar to that

Explanation:

3 0

2 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

3 years ago