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
Ivenika [448]
3 years ago
12

Please help me choose the correct statement

Chemistry
1 answer:
Pepsi [2]3 years ago
4 0
The ball has more mass because it weights more it may be less dense but more weight is being added to it
You might be interested in
Which energy resource causes the greatest waste disposal concerns?
Alex_Xolod [135]
Coal produces more pollution than any other energy source
7 0
3 years ago
When an atom gains a valence electron, it become a (n)_____ion.
Lilit [14]

An atom that gains electrons and becomes negatively charged is known as an <u>anion</u><u>.</u>

Hope it helps!

4 0
3 years ago
I JUST NEED A REAL ANSWER PLEASE.
ElenaW [278]

Explanation:

hope the picture above help u understand:)

8 0
3 years ago
Match the following aqueous solutions with the appropriate letter from the column on the right. Assume complete dissociation of
Serhud [2]

Answer:  0.17 m CH_3COONH_4 : Highest freezing point

0.20 m CoSO_4: Second lowest freezing point

0.18 m MnSO_4: Third lowest freezing point

0.42 m ethylene glycol: Lowest freezing point

Explanation:

Depression in freezing point  is a colligative property which depend upon the amount of the solute.

\Delta T_f=i\times k_f\times m

where,

\Delta T_f = change in freezing point

i= vant hoff factor

k_f = freezing point constant

m = molality

a) 0.17 m CH_3COONH_4

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i =2 for CH_3COONH_4, thus total concentration will be 0.34 m

b) 0.18 m MnSO_4

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for MnSO_4, thus total concentration will be 0.36 m

c) 0.20 m CoSO_4

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for CoSO_4, thus total concentration will be 0.40 m

d) 0.42 m ethylene glycol

For non electrolytes undergoing no dissociation, vant hoff factor is equal to 1 . Thus i = 1 for ethylene glycol, thus concentration will be 0.42 m

The more is the concentration, the highest will be depression in freezing point and thus lowest will be freezing point.

4 0
3 years ago
2. Zn + HCI. ZnCl2 + H2<br><br>what reaction is this ​
steposvetlana [31]

Answer: Type of Chemical Reaction: For this reaction we have a combination reaction. Balancing Strategies: When we add zinc to hydrochloric acid we end up with zinc chloride, a salt, and hydrogen gas. This reaction is actually a good way to make hydrogen gas in the lab.

5 0
3 years ago
Other questions:
  • The nucleus of isotope A of an element has a larger mass than isotope B of the same elemen
    7·1 answer
  • Choose the incorrect one. a. Time = Distance /Speed b. Time = Speed/Distance c. Distance = Speed x time d. Speed = Distance/time
    15·1 answer
  • 2. Explain ways friction helps us or makes things easier in our daily life. Explain the effect of friction for each. Make sure y
    10·1 answer
  • Based on what you observed, what ideas do you
    12·1 answer
  • In the manufacturing of computer chips, cylinders of silicon are cut into thin wafers that have a mass of
    12·1 answer
  • c. If the average basal rate of oxygen consumption for an adult is 15 L/hour, then calculate how many hours of oxygen use will y
    6·1 answer
  • D » » DI
    8·2 answers
  • The idea that atoms are indivisible explains all of the following except which of the following​
    15·1 answer
  • Answer now pls and i give you mega points ;)
    14·1 answer
  • Julia notices that the ""check engine"" light comes on while driving her car on days with temperatures of over 100 degrees. what
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!