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
iren2701 [21]
2 years ago
13

What happens to halogens on going top to bottom in their group

Chemistry
1 answer:
s344n2d4d5 [400]2 years ago
3 0

they remain as they are and don't change their type

You might be interested in
For the balanced chemical reaction
musickatia [10]

Answer:

150

Explanation:

  • C₄H₂OH + 6O2 → 4CO2 + 5H₂O

We can <u>find the equivalent number of O₂ molecules for 100 molecules of CO₂</u> using a <em>conversion factor containing the stoichiometric coefficients of the balanced reaction</em>, as follows:

  • 100 molecules CO₂ * \frac{6moleculesO_2}{4moleculesCO_2} = 150 molecules O₂

150 molecules of O₂ would produce 100 molecules of CO₂.

5 0
2 years ago
Which of the following elements has the highest electronegativity?
Nastasia [14]

Answer: is c

Explanation:

8 0
3 years ago
Read 2 more answers
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
Calculate the force applied to a car with mass of 1200 kg that has an acceleration of 2 m/s/s.
bearhunter [10]

Answer:

2,400

Explanation:

F = m × a

F = 1200 × 2

F = 2,400

6 0
3 years ago
Anabolic reactions _______ bonds, whereas catabolic reactions __________ bonds. A. decrease; increase. B. break; make C. weaken;
nexus9112 [7]

Answer:

The corrext answer is E. make; break

Explanation:

In living organisms, the metabolism is either anabolic or catabolic where anabolic metabolism is energy consuming and catabolic metabolism is eneegy releasesing. It should however be noted that anabolic reaction builds or biosynthesize new mollecular structures while catabolic reaction breaks down complex structure bonds into simple structures

The braking down of bonds in catabolic reations realeses energy to sustain the anabolic rection process for the formation of new bonds

6 0
3 years ago
Other questions:
  • What does the angular momentum quantum number determine?
    8·1 answer
  • A substance that will change shape to fit its container but has a definite volume is in a phase of matter?
    15·2 answers
  • Which property is not characteristic of a metal?
    12·2 answers
  • Neon is an inert gas and has the atomic number of 10. Which of the following correctly depicts its electron configuration? 1s22d
    12·1 answer
  • Which of the following is usually done last in the scientific method of investigation?
    8·2 answers
  • I am clueless on this question please help me !
    8·2 answers
  • Kinetic energy is energy an object has because of its:
    5·1 answer
  • (07.05 HC)
    11·1 answer
  • A cup of hot water loses 265 cal
    13·1 answer
  • Consider the following voltaic cell:(d) At which electrode are electrons consumed?
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!