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I am Lyosha [343]
3 years ago
6

PLEASE HELP ME!!! SCIENCE!!!

Chemistry
2 answers:
attashe74 [19]3 years ago
5 0
The correct answer is 3
sergij07 [2.7K]3 years ago
4 0

Answer:

Do Yuu Need Help???

Explanation:

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What mass of Cl2, in grams is contained in a 10.0L tank at 27oC and 3.50 atm pressure ? a. 1.42 grams b. 142 grams c. 1.01 kg d.
Evgen [1.6K]

Answer:

the correct answer is A

Explanation:

8 0
3 years ago
What is the freezing point in °C) of a 0.743 m
lesantik [10]

Answer:

-0.276 Degrees C

Explanation:

kf of water is 1.86

Freezing point depression= m x kf x i

i= ions present ( K+ Cl-) 1=2

1.86x2X.743= .276

Since its freezinf point depression the freezing point will lower

so

0-.276= -.276

7 0
3 years ago
A nuclear power plant operates at 40.0% efficiency with a continuous production of 1042 MW of usable power in 1.00 year and cons
Sergio039 [100]

Answer:

3.00 x 10^-11 joules / atom of U-235

Explanation:

We know that the formula for Power = Work done (w)/Time (t)

We need to get the joules from power , since Joules is the SI unit of work.

From the formula P = W/t

W = Power (P) * Time (t)

The SI unit for Time is seconds, hence we change 1 year in seconds

1yr * 365 days/yr * 24hrs/day * 60mins/hr * 60 secs/min = 31536000 secs

It was stated in the question that the plant operates at an efficiency of 40%,

Thus to get the true power we divide the power provided in the question by 0.4 or 40%

= X(0.4) = 1042MW

True Power X = 1042/0.4 = 2605MW

Thus true power = 2605 * 10^6 Watts

Now we have the time in seconds and true power in Watts, we then find the work done.

From our above formula P = W/t

W = P*t = (2605* 10^6) (31536000) =

Finally, we can solve for our energy (work):

P = W / T        PT = W = (2880x10^6) (31536000) = 8.22 x 10^16 joules

We then calculate the amount of energy released by only 1 single uranium-235 atom.

= 8.22 x 10^16 joules / 1.07x10^6 g U-235 (235 g / 1 mol)(1 mol/6.0210^23 atoms)

= 3.00 x 10^-11 joules / atom of U-235

5 0
3 years ago
*
Marta_Voda [28]

Answer:

proton. neutrons and electrons

5 0
3 years ago
Combustion analysis is performed on 0.50 g of a hydrocarbon, and 1.47 g of CO2 and 0.902 g of H2O are produced. What is the empi
JulijaS [17]

1. The empirical formula of the hydrocarbon is CH₃

2. The molecular formula of the hydrocarbon is C₂H₆

<h3>How to determine the mass of Carbon </h3>
  • Mass of CO₂ = 1.47 g
  • Molar mass of CO₂ = 44 g/mol
  • Molar of C = 12 g/mol
  • Mass of C =?

Mass of C = (12 / 44) × 1.47

Mass of C = 0.4 g

<h3>How to determine the mass of H</h3>
  • Mass of compound = 0.5 g
  • Mass of C = 0.4 g
  • Mass of H = ?

Mass of H = (mass of compound) – (mass of C)

Mass of H = 0.5 – 0.4

Mass of H =0.1 g

<h3>1. How to determine the empirical formula </h3>
  • C = 0.4 g
  • H = 0.1 g
  • Empirical formula =?

Divide by their molar mass

C = 0.4 / 12 = 0.03

H = 0.1 / 1 = 0.1

Divide by the smallest

C = 0.03 / 0.03 = 1

H = 0.1 / 0.03 = 3

Thus, the empirical formula of the compound is CH₃

<h3>2. How to determine the molecular formula</h3>
  • Empirical formula = CH₃
  • Molar mass = 30 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[CH₃]n = 30

[12 + (3×1)]n = 30

15n = 30

Divide both side by 15

n = 30 / 15

n = 2

Molecular formula = [CH₃]n

Molecular formula = [CH₃]₂

Molecular formula = C₂H₆

Learn more about empirical formula:

brainly.com/question/24297883

#SPJ1

4 0
2 years ago
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