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
guajiro [1.7K]
3 years ago
13

Gravitational force acts on two objects with which attribute? A. mass B. force C. orbit D. speed

Chemistry
2 answers:
Reptile [31]3 years ago
7 0
The answer is A. mass. I hope I could help. :)
Svetlanka [38]3 years ago
5 0
It is force I believe
You might be interested in
¿Cómo se llama el grupo IIA de la tabla periódica?
Verdich [7]

Answer:

los metales alcalinotérreos: berilio (Be), magnesio (Mg), calcio (Ca), estroncio (Sr), bario (Ba) y radio (Ra).

o simplemente llamado grupo 2A

Explanation:

8 0
2 years ago
How many moles contain 5.34 x 10^26 atoms of iron?
SCORPION-xisa [38]

Answer: 1.31 × 10^47 sorry if its not the answer

Explanation:

6 0
2 years ago
The charge of single electron
irga5000 [103]
The answer is B.

plz mark me as brainliest. i really need it.
8 0
3 years ago
The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a
mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g)  →  3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

3 0
3 years ago
If you initially have a volume of 4 L and a temperature of 300K. Then you decreased the volume to 2 L, what is the new temperatu
ASHA 777 [7]

Answer:

T₂ = 150 K

Explanation:

Given data:

Initial volume = 4 L

Initial temperature = 300 K

Final volume = 2 L

Final temperature = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = T₁V₂/V₁  

T₂ = 300 K × 2L / 4 L

T₂ = 600 L.K / 4 L

T₂ = 150 K

3 0
2 years ago
Other questions:
  • Which of the following metal ions is most likely to exist as a stable eight-coordinate complex with fluoride? Cu2+ Cu+ Fe3+ Ag3+
    9·1 answer
  • _______ is the current condition in the troposphere. (2 points)
    14·1 answer
  • Which of these are examples of physical changes? Check all that apply. burning wood freezing water cutting paper iron rusting mi
    5·2 answers
  • Please help with that problem <br> Important İ have quiz tomorrow
    7·1 answer
  • What is a chemical reaction that absorbs heat called?
    15·2 answers
  • Correct answer. 9. The mass number of an element is the number of? ​
    15·1 answer
  • PLS SOMONE PLS FILL OUT I WILL GIVE BRAINLIEST HURRYYY
    5·1 answer
  • Which statement best explains the purpose of using a mole in the measurement of matter? It is commonly used as another unit of m
    5·2 answers
  • Question 16 of 30
    11·1 answer
  • Acids, bases, and salts belong to a large group of compounds called:__________
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!