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slamgirl [31]
3 years ago
6

What is the mass of oxygen in 25.0 grams of potassium permanganate , KMnO4?

Chemistry
2 answers:
MArishka [77]3 years ago
4 0

Answer: 10.1 g

I just did it and got i right

Katen [24]3 years ago
3 0

Answer:

10.76 grams

Explanation:

Given that the amount of KMnO_4 is 25.0 grams.

Mass of 1 mole of KMnO_4 = 158 grams

The number of atoms in 1 mole of KMnO_4 is 4.

Mass of oxygen in 1 mole of KMnO_4 = 16\times 4 = 68 grams.

Here, 158 grams of KMnO_4 has 68 grams of oxygen

So, the amount of oxygen in 1 gram of KMnO_4 = 68/158 grams

Therefore,  the amount of oxygen in 1 gram of KMnO_4 = \frac {68}{158}\times 25 grams

=10.76 grams

Hence, 25 grams of KMnO_4 has 10.76 grams of oxygen.

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1) An unlabeled laboratory solution is found to contain a H+ ion concentration of 1 x 10-4
marishachu [46]

Answer:

C) 1 x 10-10 M

Explanation:

To solve this question we must use the equation:

Kw = [H+] [OH-]

<em>Where Kw is the equilibrium dissociation of water = 1x10-14</em>

<em>[H+] is the molar concentration of hydronium ion = 1x10-4M</em>

<em>[OH-] is the molar concentration of hydroxyl ion</em>

<em />

Replacing:

1x10-14= 1x10-4 [OH-]

<em>[OH-] = 1x10-14 / 1x10-4M</em>

<em>[OH-] = 1x10-10 M</em>

Right option is:

<h3>C) 1 x 10-10 M </h3>

7 0
3 years ago
How many liters of hydrogen gas are formed from the complete reaction of 1.03 mol of C? Assume that the hydrogen gas is collecte
aliina [53]

Answer:

27 liters of hydrogen gas will be formed

Explanation:

Step 1: Data given

Number of moles C = 1.03 moles

Pressure H2 = 1.0 atm

Temperature = 319 K

Step 2: The balanced equation

C +H20 → CO + H2

Step 3: Calculate moles H2

For 1 mol C we need 1 mol H2O to produce 1 mol CO an 1 mol H2

For 1.03 moles C we'll have 1.03 moles H2

Step 4: Calculate volume H2

p*V = n*R*T

⇒with p = the pressure of the H2 gas = 1.0 atm

⇒with V = the volume of H2 gas = TO BE DETERMINED

⇒with n = the number of moles H2 gas = 1.03 moles

⇒with R = the gas constant = 0.08206 L*Atm/mol*K

⇒with T = the temperature = 319 K

V = (n*R*T)/p

V = (1.03 * 0.08206 *319) / 1

V = 27 L

27 liters of hydrogen gas will be formed

8 0
3 years ago
What is the formula for barium iodide?
zysi [14]
Formula :  BaI₂. <span>barium iodide</span>
6 0
3 years ago
Sugar dissolves readily in water because it is a(n) ____ substance.
Mnenie [13.5K]
The answer to this question would be D. Hydrophilic.

The word hydrophilic mean attracted by water. That means the molecule has a force to attract water molecule, thus be able to dissolve in water.  The polarity of the molecule would determine whether a molecule hydrophilic or not.
Its opposite would be hydrophobic which the molecule can't dissolve in water.  One example of this would be oil or fat. That is why sometimes it is called lipophilic too.
3 0
3 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
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