A simple way to go about this is that we look at the solubility curve, on the x axis we first look at the temperature and then the corresponding value of solute/100g H2O on the y axis, from the 4 curves above only NaNO3 has a curve that can accommodate 80g of salt at 40 without being Saturated since at 40 degrees it can accommodate 105g of salt to become completely Saturated.

7 0

2 years ago

Read 2 more answers

PLEASE ANSWER CORRECTLY FOLLOWING GUIDLINES SO ANSWER WONT GET DELETED I REALLY NEED HELP

blagie [28]

Chemical reaction

equation

reactants

products

yields

mass

balanced

atoms

coefficients

numbers

element

correct

substance

two

Explanation:

did my best lol I'm like 98.69% confident

8 0

2 years ago

0.785 moles of N2,fill a balloon at 1.5 atm and 301 K.<br> What is the volume of the balloon?

Tcecarenko [31]

Answer:

V = 12.93 L

Explanation:

Given data:

Number of moles = 0.785 mol

Pressure of balloon = 1.5 atm

Temperature = 301 K

Volume of balloon = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

Now we will put the values.

V = nRT/P

V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm

V = 19.4 L /1.5

V = 12.93 L

7 0

2 years ago

Calculate the energy, in joules, required to ionize a hydrogen atom when its electron is initially in the n =2 energy level. The

qaws [65]

Answer:

$E_{ionization}=5.45\times 10^{-19}\ J$

Explanation:

$E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

So, $n_i=2$ and $n_f=\infty$ (As the hydrogen has to ionize)

Thus,

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2} - \dfrac{1}{{\infty}^2})\ J$

$\Delta E=2.18\times 10^{-18}(\frac{1}{2^2})\ J$

$E_{ionization}=5.45\times 10^{-19}\ J$

4 0

3 years ago