Match the standard reduction potentials for each half reaction and the electric potential for the voltaic cell.

A meteorologist filled a weather balloon with 3.00L of the inert noble gas helium. The balloon's pressure was 765 torr. The ball

yanalaym [24]

Answer:

4.33 L

Explanation:

Step 1: Given data

Initial volume of the balloon (V₁): 3.00 L

Initial pressure of the balloon (P₁): 765 torr

Final volume of the balloon (V₂): ?

Final pressure of the balloon (P₂): 530 torr

Step 2: Calculate the final volume of the balloon

If we consider Helium to behave as an ideal gas, we can calculate the final volume of the balloon using Boyle's law.

$P_1 \times V_1 = P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{765torr \times 3.00L}{530torr} = 4.33 L$

7 0

3 years ago

Find the new pressure of 5 L of a gas at STP that is cooled to 250 K and expanded to 6L

UNO [17]

I’m so sorry i have to do this 5L 24

5 0

3 years ago

What was the old percentages of copper to zinc before 1983?

vovikov84 [41]

Answer:

95% Copper and 5% Zinc

Explanation:

Easy stuff read in the chemistry book...

3 0

3 years ago

What is a substance and what are the two types of substances?

masha68 [24]

A pure substance has a constant composition and cannot be separated into simpler substances by physical means. There are two types of pure substances: elements and compounds. Elements: are pure substances made up of only l type of atom. Atoms of the same element are identical in properties.

6 0

3 years ago

Read 2 more answers

Give regents and mechanism

Kryger [21]

Answer:

Reagents: 1) $BH_3$ 2) $H_2_O2$ , $OH^-$

Mechanism: Hydroboration

Explanation:

In this case, we have a hydration of alkenes reaction. But, in this example, we have an anti-Markovnikov reaction. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: "hydroboration".

The first step of this reaction is the addition of borane ( $BH_3$ ) to the double bond. Then in the second step, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the third step, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond " $BH_2-O-OH$ ". In step number 4 we have the migration of the C-B bond to oxygen. Then in step number 5, we have the attack of $OH^-$ on the $O-BH_2$ to produce an alkoxide. Finally, the water molecule produce in step 2 will protonate the molecule to produce the alcohol.

See figure 1

I hope it helps!

4 0

3 years ago