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KengaRu [80]
3 years ago
12

What mass, in grams, of oxygen gas (o2) is contained in a 8.5-liter tank at 32.0 degrees celsius and 3.23 atmospheres?

Chemistry
1 answer:
e-lub [12.9K]3 years ago
5 0
For this problem we assume that oxygen is an ideal gas. So, we use the equation PV=nRT where P is pressure, V is the volume, n is the number of moles, R is a universal constant and T is the temperature. We first solve for the number of moles n. Then, using the molar mass of oxygen we convert it to grams. 

PV=nRT
n = PV / RT
n = 3.23 (8.5) / 0.08206 (32+273.15)
n = 1.0964 mol 

mass = 1.0964 mol (32g / 1 mol) = 35.09 g O2
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The alkali earth metal beryllium (Be) engages in a chemical reaction and loses all of its valence electrons.
denpristay [2]

The loss of electron from an results in the formation of cation represented by the positive charge on the element whereas gaining of electron results in the formation of anion represented by the negative charge on the element.

The alkali earth metal beryllium (Be) belongs to the second group of the periodic table. The ground state electronic configuration of Be is:1s^{2}2s^{2}

From the electronic configuration it is clear that it has 2 valence electrons in its valence shell (2s^{2}).

After losing all valence electrons that is 2 electrons from 2s orbital. The electronic configuration will be:

1s^{2}2s^{0}

Since, lose of electron is represented by positive charge on the element symbol. So, the beryllium will have +2 charge on its symbol as Be^{2+}.

Hence, beryllium will have 2+ charge on it after losing all its valence electrons in the chemical reaction.

6 0
3 years ago
Read 2 more answers
what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?
MaRussiya [10]

Answer:

Explanation:

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now  F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻  +  H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF,  the [ OH ⁻] from the given pH, and finally the mass needed to produce that  OH⁻ concentration.  

The equilibrium constant, Kb , can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is  6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14  ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information :

                                   F⁻                    HF                        OH⁻

Equilibrium                 X                  2.51 x 10⁻⁶            2.51 x 10⁻⁶

(2.51 x 10⁻⁶)² / X  =  1.5 x 10⁻¹¹     ⇒  X =  (2.51 x 10⁻⁶)²  / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL ( 0.35 L ) we need to add:

0.41 mol HF/ 1 L  *  0.35 L = 0.144 mol

and finally the mass will be:

0.144 mol NaF *  42.0 g/mol NaF = 6.03 g NaF

7 0
2 years ago
What do engineers use to test their designs of new technologies?
erica [24]
The answer Fam is B) Models
4 0
3 years ago
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Manganese-56 is a beta emitter with a half-life of 2.6 hours. What is the mass of
Arisa [49]

Answer:

0.0009765625

Explanation: u just keep doing a half life till it goes to 13 aka

8.0g divided by 2  till you reach to 13 times of a half life ;p

8 0
2 years ago
Read 2 more answers
A student uses a solution of potassium hydroxide to titrate a solution of nitric acid. Which question is the student trying to a
Mnenie [13.5K]

Answer:

A. What is the concentration of nitric acid?

Explanation:

its right

4 0
2 years ago
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