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castortr0y [4]
3 years ago
6

Three chemistry students measured the length of a copper bar. The recorded lengths were 5.05 cm, 5 cm , and 5.1 cm, What is the

average length of the bar by showing my work ?
Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
3 0
5.05 + 5 + 5.1 = 15.15cm Then you just divide it by the amount of measurements you had like this:15.15 ÷ 3 = 5.04999971cm Then you can just round it to the 3rd figure: 5.05cm < And that's the mean/average length of the bar. :) (Or the one above if you want all of the decimals too) 
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On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?
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Answer:

The answer is "2%"

Explanation:

Equation:

HNO_2\ (aq) \leftrightharpoons  H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\  K_a = 4.0\times \ 10^{-4}

H^{+}=?

Formula:

Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}

Let

[H^{+}] = [NO_2^{-}] = x at equilibrium

x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \  M\\\\

therefore,

[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M

Calculating the % ionization:

= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\

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Combining hydrogen and oxygen to make water Group of answer choices releases a lot of energy. consumes a lot of energy. releases
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Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation:
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Answer:

See explanation

Explanation:

First, let's write the balanced equation again:

2 NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)

Now, we know that the total pressure was 7.76 atm. This total pressure, is the sum of the pressure of water and CO2 like this:

Ptotal = Pwat + PCO2 (1)

This is the dalton's law for partial pressures.

The pressure can be also be relationed with the moles

Ratio of mole = Ratio of pressure

so, taking this in consideration we can say the following:

Pwater/PCO2 = moles water / moles CO2

As the only components exerting pressure are CO2 and Water (Because they are in gas phase), the total pressure can be splitted between the two of them so:

Pwater = Ptotal/2

Pwater = 7.76 / 2 = 3.88 atm

With this pressure, and using the ideal gas equation, we can know the moles of water:

PV = nRT

n = PV/RT     using R = 0.082 L atm / K mol

n = 3.88 * 5 / 0.082 * (160+273)

n = 0.546 moles of water

b) now that we have the moles of water, we can actually know the moles that reacted originally from the sodium carbonate by stechiometry.

2NaHCO3(s) <-> Na2CO3(s) + H2O(g) + CO2(g)    MMCO2 = 84 g/mol

the moles of NaHCO3 initially:

n = 100 / 84

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so, If 1.19 moles of NaHCO3 reacted, and only produces 0.546 moles of water and CO2, then, the remaining moles of NaHCO3 is:

remaining moles = 1.19 - 0.546 = 0.644 moles

therefore the mass remaining:

mCO2 = 0.644 * 84

mCO2 = 54.096 g

c) As it was stated before, only the gaseous components are involved in the pressure, thus, in the kp expression which is:

Kp = Pwater * PCO2

Kp = 3.88 * 3.88

Kp = 15.0544

d) As the total pressure is 7.76 atm and the fact that NaHCO3 is solid, this component is not exerting any pressure in the reaction, as seen in the Kp expression, so it won't matter that if we raise a little the quantity of the reactant, it still has some remaining.

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3 years ago
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