The electrons that are passed to nadp at the end of the electron transfer chain were obtained from

How many grams of oxygen gas (02) are needed to completely react with 9.30 moles

Luba_88 [7]

Answer:

223 g O₂

Explanation:

To find the mass of oxygen gas needed, you need to (1) convert moles Al to moles O₂ (via the mole-to-mole ratio from reaction coefficients) and then (2) convert moles O₂ to grams O₂ (via the molar mass). When writing your ratios/conversions, the desired unit should be in the numerator in order to allow for the cancellation of the previous unit. The final answer should have 3 sig figs because the given value (9.30 moles) has 3 sig figs.

4 Al + 3 O₂ ----> 2 Al₂O₃
^ ^

Molar Mass (O₂): 32.0 g/mol

9.3 moles Al 3 moles O₂ 32.0 g
------------------- x --------------------- x -------------------- = 223 g O₂
4 moles Al 1 mole

6 0

2 years ago

If a proton and an electron are released when they are 5.50×10−10 mm apart (typical atomic distances), find the initial accelera

Vlada [557]

Answer: The initial acceleration of the proton = (4.56 × 10^23) m/s2

The initial acceleration of the electron = (8.36 × 10^26) m/s2

Explanation: The force of attraction between the proton and electron can be computed using the statements of Coulomb's law which state that the force of attraction between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of their distances apart.

F = (Kq1q2)/(r^2) where K = (9 × (10^9) Nm(C^-2))

But q1 is the charge on a proton = (1.6 × (10^-19)) C

q2 is charge on an electron = -(1.6 × (10^-19)) C

r = (5.50 × (10^-10))mm = (5.50 × (10^-13))m

Computing all that, F = 0.0007616529 N = (7.62 × 10^-4) N

But the force of attraction is converted to that required for motion when they're released.

F = ma.

For proton, m = (1.67 × 10^-27) kg

a = F/m = 0.000762/(1.67 × 10^-27) = (4.56 × 10^23) m/s2

For electron, m = (9.11 × 10^-31) kg

a = F/m = 0.000762/(9.11 × 10^-31) = (8.36 × 10^26) m/s2

QED!

7 0

3 years ago

A gas occupies 4.30 Liters at STP. What volume does it occupy at 2.56 atm and 302

laiz [17]

Answer:

V₂ = 1.86 L

Explanation:

Given data:

Initial volume = 4.30 L

Initial pressure = 1 atm

Initial temperature = 273.15 K

Final temperature = 302 K

Final volume = ?

Final pressure = 2.56 atm

Solution:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂ /T₁ P₂

V₂ = 1 atm ×4.30 L × 302 K / 273.15 K × 2.56 atm

V₂ = 1298.6 atm.L.K / 699.26 K.atm

V₂ = 1.86 L

4 0

3 years ago

2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme

mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

$2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O$

Next, we assign the oxidation numbers as follows:

$2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}$

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

4 0

3 years ago

which of the following is a balanced chemical equation for the reaction of magnesium with nitrogen gas to form magnesium nitride

ioda

So the unbalanced equation would be Mg + N^2 --> Mg^3N^2

Which means the balanced equation would be 3Mg + N^2 --> Mg^3N^2

This is balance the equation out since you now has 3 magnesium and 2 nitrogen on the left side, and 3 magnesium on 2 nitrogen on the right. Double check my work though, it's been awhile.

8 0

4 years ago