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diamong [38]
1 year ago
8

P4O10 + H2O —> what are the products

Chemistry
1 answer:
liubo4ka [24]1 year ago
3 0

Answer:

P4O10 + H2O → H3PO4

Explanation:

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Rank the following measurements in order from least to greatest
Harman [31]
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1.A
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4 0
3 years ago
What is the volume, in liters, of 1.40 mol of oxygen gas at 20.0°C and 0.974 atm?
topjm [15]

Answer:

V = 34.55 L

Explanation:

Given that,

No of moles, n = 1.4

Temperature, T = 20°C = 20 + 273 = 293 K

Pressure, P = 0.974 atm

We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :

PV=nRT

R is gas constant, R=0.08206\ L-atm/mol-K

Finding for V,

V=\dfrac{nRT}{P}\\\\V=\dfrac{1.4\times 0.08206\times 293}{0.974 }\\\\V=34.55\ L

So, the volume of the gas is 34.55 L.

4 0
3 years ago
A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was dil
Masteriza [31]

Answer : The value of pK_a of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}

\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol

Now we have to calculate the value of pK_a of the weak acid.

The equilibrium chemical reaction is:

                          HA+KOH\rightleftharpoons HK+H_2O

Initial moles     0.25     0.03        0

At eqm.    (0.25-0.03)   0.03      0.03

                     = 0.22

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[HK]}{[HA]}

Now put all the given values in this expression, we get:

3.85=pK_a+\log (\frac{0.03}{0.22})

pK_a=4.72

Therefore, the value of pK_a of the weak acid is, 4.72

7 0
3 years ago
Please help me with this question it's due in like 3hours from now
saw5 [17]

237kj mole for both.....




4 0
3 years ago
Can anyone help me? thank you :))
Alex_Xolod [135]
8. 63360
9. 1000
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