Answer:
1.2*10^24 molecules of CF4
Explanation:
the molar mass of cf4 is 88.0043 g/mol
176/88.0043 = 2 moles of CF4
Then multiply by avogadro's number (6.022*10^23) to get the number of molecules
2*6.022*10^23 = 1.2*10^24 molecules of CF4
The total Pressure equals the sum of all pressures contained
<span>Since total pressure and the pressure of nitrogen and oxygen is given, finding the pressure of carbon dioxide is given by: </span>
<span>Pressure of Carbon dioxide = 42.9- 6.6- 23.0 </span>
<span>=13.3kPa </span>
- The student weighs out 0.0422 grams of the metal magnesium, thus we can figure that the more's, the magnesium he used, is the mass of the magnesium over the more mass, which is 0.024422.
- That is approximately 0.001758.
- Furthermore, it claims that too much hydrochloric acid causes the metal magnesium to react, producing hydrogen gas.
- The volume of collected gas is 43.9 cc, the mastic pressure is 22 cc, and a sample of hydrogen gas is collected over water in a meter.
<h3>Is it true that calculations made utilizing experimental and gathered data result in a percent error? </h3>
- Consequently, we are aware that magnesium and chloride react.
- We create 1 as the reaction ratio is 1:2.
- The hydrogen and 1 are more.
- Magnesium chloride is more.
- Therefore, based on this equation, we can infer that the amount of hydrogen that would be created in this scenario is greater than the amount of magnesium present here, or 0.001758 more.
- Among hydrogen, there is.
- \Once we convert the temperature from 32 Celsius to kelvin, we can tell you that the temperature is actually about 5.15 kelvin.
- The gas has a volume of 43 in m, which is equal to 0.0439 liter and indicates that the pressure is approximately 832 millimeter.
- Mercury, which is 2 times 13332 plus ca, or roughly 110922.24 par, is a mathematical constant.
- So, in this instance, we are aware that p v = n r t.
- The r in this case equals p v over n t, thus we want to determine the r.
- So p is 110922.24. The temperature is 305.15 and the V is 0.04 over the n is 0.001758.
- Let's proceed with the calculations right now.
- In this instance, you will discover that the solution is 9.077 times 10; that is all there is to it.
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Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
Answer:
C) 712 KJ/mol
Explanation:
- ΔH°r = Σ Eb broken - Σ Eb formed
- 1/2Br2(g) + 3/2F2(g) → BrF3(g)
∴ ΔH°r = - 384 KJ/mol
∴ Br2 Eb = 193 KJ/mol
∴ F2 Eb = 154 KJ/mol
⇒ Σ Eb broken = (1/2)(Br-Br) + (3/2)(F-F)
⇒ Σ Eb broken = (1/2)(193 KJ/mol) + (3/2)(154 KJ/mol) = 327.5 KJ/mol
∴ Eb formed: Br-F
⇒ Σ Eb formed (Br-F) = Σ Eb broken - ΔH°r
⇒ Eb (Br-F) = 327.5 KJ/mol - ( - 384 KJ/mol )
⇒ Eb Br-F = 327.5 KJ/mol + 384 KJ/mol = 711.5 KJ/mol ≅ 712 KJ/mol
