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2 years ago

7

An experiment produced 0.10 g CO2 with a volume of 0.056 L at STP. If the accepted density of CO2 at STP is 1.96 g/L, what is th

e approximate percent error?

1 answer:

Mandarinka [93]2 years ago

3 0

The experimental density of CO2 at STP is 0.10/0.056=1.78 g/L. The percent error equals to (1.96-1.78)/1.96*100%=9.18%. So the answer is 9.18%.

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Find the ph of a 0.250 m solution of nac2h3o2. (the ka value of hc2h3o2 is 1.80×10−5). express your answer numerically to four s

Slav-nsk [51]

Answer is: pH value of solution of NaC₂H₃O₂ is 9.07.
Chemical reaction: C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻.
Ka(HC₂H₃O₂) = 1,8·10⁻⁵.<span>
Ka · Kb = Kw.
</span>1,8·10⁻⁵ mol/dm³ · Kb = 1·10⁻¹⁴ mol²/dm⁶; the ionic product of water at 25°C.<span>
Kb(</span>C₂H₃O₂⁻) = 1·10⁻¹⁴ mol²/dm⁶ ÷ 1,8·10⁻⁵ mol/dm³.<span>
Kb(</span>C₂H₃O₂⁻) = 5,56·10⁻¹⁰ mol/dm³.
c(C₂H₃O₂⁻) = 0,25 M.
[OH⁻] = [HC₂H₃O₂] = x.
[C₂H₃O₂⁻] = 0,25 M - x.
Kb = [OH⁻] · [HC₂H₃O₂] / [C₂H₃O₂⁻].
5,56·10⁻¹⁰ = x² / (0,25 M -x).
Solve quadratic equation: x = [OH⁻] = 0,0000118 M.
pOH = -log[OH⁻] = -log(0,0000118M) = 4,93.
pH + pOH = 14.
pH = 14 - 4,93 = 9,07.

6 0

3 years ago

A compound is found to contain 43.66% of P and 56.33% of O.

jek_recluse [69]

Answer: P2O5 is the empirical formula.

Explanation: When given percentages you can assume that many grams of each atom are in the compound. Then you divide grams by the molar mass of each element, giving you moles. Once you have moles, divide by the smaller molar amount, which should give you 1 mol of Phosphorus and 2.5 mol of Oxygen. Then multiply by 2 in order for both moles to be a whole number. This gets you 2 and 5.

6 0

2 years ago

If the volume of wet gas collected over water is 85.0 mL at 20°C and 760 mm Hg , what is the volume of dry gas at STP conditions

dimulka [17.4K]

Answer: 77.4 mL

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg

$P_2$ = final pressure of dry gas at STP = 760 mm Hg

$V_1$ = initial volume of dry gas = 85.0 mL

$V_2$ = final volume of dry gas at STP = ?

$T_1$ = initial temperature of dry gas = $20^oC=273+20=293K$

$T_2$ = final temperature of dry gas at STP = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get the final volume of wet gas at STP

$\frac{742.5mmHg\times 85.0ml}{293K}=\frac{760mmHg\times V_2}{273K}$

$V_2=77.4mL$

Volume of dry gas at STP is 77.4 mL.

5 0

3 years ago

11. If four gases in a cylinder each exert 6 atm, what is the total pressure exerted by the

Debora [2.8K]

Answer:

24 atm is the total pressure exerted by the gases

Explanation:

We propose this situation:

In a vessel, we have 4 gases (for example, hydrogen, Xe, methane and chlorine)

Each of the gases has the same pressure:

6 atm → hydrogen

6 atm → xenon

6 atm → methane

6 atm → chlorine

To determine the total pressure, we sum all of them:

Partial pressure H₂ + Partial pressure Xe + Partial pressure CH₄ + Partial pressure Cl₂ = Total P

6 atm + 6 atm + 6 atm + 6 atm = 24atm

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3 years ago

The electronic configuration of an element is given below.

Amiraneli [1.4K]

The statement which is true about the reactivity of element with 1S²2S²2P⁶3S¹ is

it is reactive because it has to lose one electron to have a full outermost energy level.

<u><em>Explanation</em></u>

<u><em> </em></u>Element with 1S²2S²2P⁶3S¹ electron configuration is a sodium metal.

sodium has one electron in the outermost energy level.

for sodium to have a full outermost energy level ( 8 electrons) it loses the 1 electron in 3S¹ to form a positively charged ion. (Na⁺)

4 0

2 years ago