What is the mass percent of water in copper (II) sulfate?

Five gases combined in a gas cylinder have the following partial pressures: 3.00 atm (N2), 1.80 atm (O2), 0.29 atm (Ar), 0.18 at

levacccp [35]

In order to solve the total pressure that is exerted by the gases, we need to use the Dalton's Law of Partial pressures. These are the calculations that you need to find out the total amount of pressure exerted to the gases:

3.00atm (N2) + 1.80atm (O2) + 0.29atm (Ar) + 0.18atm (He) + 0.10atm (H),
add up all of that, and the answer would turn out to be: 5.37atm.

8 0

2 years ago

Read 2 more answers

Calculate the empirical formula of a compound that has a composition of 5.9% (by mass) hydrogen and 94.1% (by mass) oxygen.

adelina 88 [10]

Answer:

The empirical formula is the simplest form;

Given:

Oxygen O at 94.1% and

H at 5.9%

Assume 100grams.

94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O

5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H

There is one mole of O for each mole of H so the empirical formula is $O_1H_1$

and written as OH.

8 0

2 years ago

Read 2 more answers

How many kilojoules are released when 8.2 g of water condenses at 100 °c and cools to 15 °c?

MrRa [10]

Answer:- 2.92 kJ of heat is released.

Solution:- We have water at 100 degree C and it's going to be cool to 15 degree C.

So, change in temperature, $\Delta T$ = 15 - 100 = -85 degree C

mass of water, m = 8.2 g

specific heat of water, c = $4.184\frac{J}{^0C.g}$

The equation used for solving this type of problems is:

$q=mc\Delta T$

Let's plug in the values in the equation and solve it for q which is the heat energy:

q = (8.2)(4.184)(-85)

q = -2916.248 J

They want answer in kJ. So, let's convert J to kJ and for this we divide by 1000.

$q=-2916.248J(\frac{1kJ}{1000J})$

q = -2.92 kJ

Negative sign indicates the heat is released. So, in the above process of coiling of water, 2.92 kJ of heat is released.

3 0

3 years ago

How many atoms of carbon are represented in 3CH3CH20

pishuonlain [190]

6 carbon atoms

          H H
         | |
3 x   H - C - C - O - H
   | |
      H H

8 0

2 years ago

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

4 0

3 years ago