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NeX [460]
3 years ago
8

How many grams are contained in a 0.183 mol sample of ammonium phosphate?

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
7 0

Molar mass of ( NH₄)₃PO₄ = 14.01×3 + 1.01×12 + 30.97 + 16.00×4 = 149.12 g/mol. Mass of 0.183 mol ...

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How many gallons of gasoline that is 5% ethanol must be added to 2,000 gallons of gasoline with no ethanol to get a mixture that
Amanda [17]

Answer:

V_1= 3000 gal

Explanation:

We have 3 solutions:

  • Solution 1 (with ethanol)
  • Solution 2 (no ethanol)
  • Final solution

V_f=V_1 + V_2

and for the ethanol:

V_f*0.03=V_1*0.05 + V_2*0

V_f=V_1 \frac{5}{3}

Combining:

V_1 \frac{5}{3}=V_1 + V_2

V_1 \frac{2}{3}= V_2

V_1= \frac{3}{2} V_2

If V2=2000 gal:

V_1= \frac{3}{2} 2000 gal

V_1= 3000 gal

8 0
3 years ago
Enter an equation to show how h2po3− can act as a base with hs− acting as an acid. Express your answer as a chemical equation. I
vladimir2022 [97]

The acid - base equation between H2PO3^- and HS^- is H2PO3^- +  HS^- ⇄S^- + H3PO3.

<h3>What is an acid?</h3>

An acid is a substance that can donate hydrogen ions while a base is a substance that can accept hydrogen ion. This is the acid base definition according to Brownstead - Lowry.

To show the acid - base relationship between H2PO3^- and HS^-, we have the equation;

H2PO3^- +  HS^- ⇄S^- + H3PO3

Learn more about acids and bases: brainly.com/question/10282816

4 0
2 years ago
The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate
Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

6 0
3 years ago
What are equation for the chemical change that produces water from two hydrogen molecules and one oxygen molecule. And the react
aliya0001 [1]

Answer:

h2+O ---> H2O

reactants: H2 & O

products: H2O

Explanation:

The simple reaction that produces a water molecule from H2 and O would be the one written above, even though there are 2 hydrogen molecules, they will form an H2 molecule rather than 2 individual H molecules (almost never seen) the reactants would be your hydrogen and oxygen molecules individually before they bond to form a molecule of water (H2O) which is the product

6 0
2 years ago
15.How can a pure sample of barium sulphate be otained from barium carbonate?
yan [13]

Answer:

Barium carbonate powder is stirred add pulp in the entry, the vitriol that the adds solubility then reaction that makes the transition is filtered and is obtained the barium sulfate filter cake and liquid after the transition.

8 0
3 years ago
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