1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ierofanga [76]
3 years ago
6

PLEASE HELP!

Chemistry
1 answer:
Scilla [17]3 years ago
8 0

Answer:

P₂ = 450 kiloPascals

Explanation:

Boyle's law :))

P₁V₁ = P₂V₂

300*75 = P₂*50

P₂*50= 300*75

P₂ = 300*75/50 = 450

P₂ = 450 kiloPascals

<em>The weight has expanded because of pressure of gas.</em>

You might be interested in
The nature of zinc powder and Cobalt (II )oxide is heated the following reaction occurs ;
Scilla [17]

1) The metal which reduces the other compound is the one higher in the reactivity. So in this case it is \mathrm{Zn}.

2) The substance which brings about reduction while itself getting oxidised (that is losing electrons) is called a reducing agent. Here, $\mathrm{Zn}$ is the reducing agent and reduces Cobalt Oxide to Cobalt while itself getting oxidised to Zinc oxide.

8 0
3 years ago
Please refer to image please
GarryVolchara [31]

Answer:

76.9L

Explanation:

Based on the graph, whenever the temperature increases by 100K, the volume increases by 10L, so do 769/10

7 0
2 years ago
I need help with 4, 5, 8, 9, and 6. Quickly I need it before class starts. Worth points!!!!!! HelP
GalinKa [24]

Answer:

4. 264.6J

5. 37.5J

6. 96J

7. 55Watts

8. 77.14m

9. 6s

10. 750Watts

Explanation:

4). Mechanical energy (potential energy) = mass (m) × acceleration due to gravity (g) × height (h)

m = 3kg, h = 9m, g = 9.8m/s²

P.E = 3 × 9 × 9.8

= 264.6J

5). Kinetic energy (K.E) = 1/2 × m × v²

Where;

m = mass (kg) = 3kg

v = velocity (m/s) = 5m/s

K.E = 1/2 × 3 × 5²

K.E = 1/2 × 3 × 25

K.E = 1/2 × 75

K.E = 37.5J

6). Work done (J) = Force (N) × distance (m)

Force = 12N, distance = 8m

Work done = 12 × 8

= 96J

7). Power = work done (J) ÷ time (s)

Work done = 550J, time = 10s

Power = 550/10

= 55Watts.

8). Work done = force (F) × distance (m)

Work done = 540J, force = 7N, distance = ?

540 = 7 × d

540 = 7d

d = 540/7

d = 77.14m

9). Power = work done (J) ÷ time (s)

Work done = 300J, time = ?, Power = 50Watts.

50 = 300/t

50t = 300

t = 300/50

t = 6seconds.

10). Power = work done (J) ÷ time (s)

This means that;

Power = force × distance / time

Force = 300N, distance = 5m, time = 2s

Power = 300 × 5 ÷ 2

Power = 1500 ÷ 2

Power = 750Watts

3 0
3 years ago
Is this statement true or false? Bases feel slippery. A. True B. False
Ksenya-84 [330]
Actually, it is true

7 0
3 years ago
Read 2 more answers
What element has the electron configuration below? *<br><br> please help me
timofeeve [1]

Answer:

Symbol Ar

Group 18

Electron configuration- 1s² 2s² 3p6 3s² 3p6

Explanation:

The 6 is small and will be placed in top but I don't have the option that's why I wrote like that

8 0
2 years ago
Other questions:
  • A liquid is considered an acid if it has a lot of
    8·1 answer
  • Which of these processes involve the weakening of the attraction between particles
    5·1 answer
  • Dehydration of 2-methyl-2-pentanol forms one major and one minor organic product. Draw the structures of the two organic product
    7·1 answer
  • 2C6H14 + 19O2 → 12CO2 + 14H2O How many atoms, total, are reacting in this equation?
    12·2 answers
  • HClO4 acid solution has a concentration of 5 Molarity. Calculate the concentration of this solution in
    13·1 answer
  • What is the solute in a solution
    10·2 answers
  • Which is the best way to turn helium gas into liquid helium?
    13·1 answer
  • PLEASE HELP ME.
    6·2 answers
  • Need help I don’t understand this at all need to show my work and strategy it’s stoichiometry gram to gram please help
    10·1 answer
  • Uno de los componentes del “sal de uvas”, es el bicarbonato de sodio, es un sólido cristalino blanco, debido a
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!