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4 years ago

Electrons spill in shells in the nucleous the closest shell (n=1) can contain the maximum of. Electrons

Chemistry

1 answer:

stiks02 [169]4 years ago

8 0

The answer is 2 electrons I believe!

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The velocity (speed) of an object was determined to be 45 miles per

Ahat [919]

Answer: 72.58 km/hr

Explanation:

$\frac{45 miles}{1 hour}$ × $\frac{1 km}{0.62 miles}$ = 72.58 km/hr

4 0

3 years ago

Read 2 more answers

How many moles of electrons are required to reduce one mole of MnO4- to one mole of Mn2+? (2 points)

Dafna1 [17]

The answer would be 6 e-. This is becuase you are turning a charge of -4 into a +2. In order to do this, you transfer 4 electrons for a neutral charge, and an additional 2 for a charge of +2.

This makes a total charge of +2, and the total transferred electrons 6 e-

8 0

3 years ago

Help pls match allll

AleksAgata [21]

2-A
1-B
5-C
4-D
3-E

I hope this helped:)

8 0

3 years ago

PLEASE HELP!!!!!!

pogonyaev

Answer:

$m_{B}^{theoretical}=0.365gB$

$Y=87.1\%$

Explanation:

Hello there!

In this case, since the reaction (A->B) have an initial amount of pure 4-aminobenzoic acid, the first step to compute the theoretical yield is to solve the following stoichiometric setup:

$m_{B}^{theoretical}=0.303gA*\frac{1molA}{137.14gA}*\frac{1molB}{1molA}*\frac{165.19 gB}{1molB}\\\\ m_{B}^{theoretical}=0.365gB$

Whereas A stands for 4-aminobenzoic acid and B for the benzocaine. Moreover, we compute the percent yield by dividing the actual yield (0.318 g) by the theoretical one (0.365 g):

$Y=\frac{0.318g}{0.365g} *100\%\\\\Y=87.1\%$

Best regards!

4 0

3 years ago

Predict the boiling point of water at a pressure of 1.5 atm.

Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

$ln\frac{P_{1} }{P_{2}} =$ -Δ $\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )$

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

$T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}$

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

7 0

3 years ago