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
sertanlavr [38]
2 years ago
6

Help please help me!!!!!

Chemistry
1 answer:
BlackZzzverrR [31]2 years ago
3 0

Answer:

Maybe the no A is correct or c

You might be interested in
What are the charges of families 1, 2, & 13 (boron family) if/when they become Ions
Natalka [10]

Answer:

Explanation:

Group one elements are alkali metals. All alkali metal have one valance electron. They loses their one valance electron and from cation with charge of +1.

Charges on group one.

Hydrogen = +1

Lithium = +1

Sodium = +1

Potassium = +1

Rubidium = +1

Cesium = +1

Francium = +1

Group two elements are alkaline earth metals. All alkaline earth metal have two valance electron. They loses their two valance electron and from cation with charge of +2.

Charges on group two.

Beryllium = +2

Magnesium = +2

Calcium = +2

Strontium = +2

Barium= +2

Radium = +2

Group 13 elements are boron family. All elements have three valance electrons. They loses their three valance electron and from cation with charge of +3.

Charges on group 13.

Boron = +3

Aluminium = +3

Gallium = +3

Indium = +3

Thallium= +3

Group 13 elements are also shows +1 charge by losing one valance electron.

8 0
3 years ago
Determine which is the control group and which is the experimental group in the following scenario ( Scientists are testing a ne
Daniel [21]

Answer:

The group given the sugar pills.

Explanation:

The control group is the ones given sugar pills because they did not get the experimental aspirin

3 0
2 years ago
Read 2 more answers
Given the balanced equation: 2KClO₃ ---> 2KCl + 3O₂ How many moles of O₂ are produced when 4.0 moles of KCl are produced?
Verizon [17]

Answer:

The answer to your question is 6.0 moles of O₂

Explanation:

Data

                      2KClO₃    ⇒     2KCl    +    3O₂

moles of O₂ = ?

moles of KCl = 4

Process

To find the number of moles of O₂, use proportions and cross multiplication.

Use the coefficients of the balanced equation.

                    2 moles of KCl ----------------- 3 moles of O₂

                    4 moles of KCl -----------------  x

                          x = (4 x 3) / 2

-Simplification

                          x = 12/2

-Result

                        x = 6 moles of O₂

-Conclusion

When 4,0 moles of KCl are produced, 6.0 moles of O₂ will be produced.                          

6 0
2 years ago
4.7C Renewable and Nonrenewable Resources
ki77a [65]
Answer: petroleum
because it is a fossil fuel like coal, and natural gas.
4 0
2 years ago
The chemical formula for disulfur monoxide is:<br> A. S2O<br> B. 2SO<br> C. SO2<br> D. S2O2
Delvig [45]
The answer is (A).
Hope this helps :).

3 0
2 years ago
Other questions:
  • Help?
    5·2 answers
  • Which of the following is an incorrect way of rearranging the density formula:
    8·2 answers
  • Which description is not a property of an acid? A turns litmus paper red B bitter taste C corrosive D dissolve metals
    12·1 answer
  • How many oxygen atoms are present in the compound aluminum sulfate?
    5·1 answer
  • 1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?
    9·1 answer
  • Which statement is TRUE
    12·2 answers
  • Iron(II) chloride and sodium carbonate react to make iron(II) carbonate and sodium chloride: FeCl2(aq) + Na2CO3(s) → FeCO3(s) +
    15·2 answers
  • If 16.4 grams of calcium nitrate is heated as shown in the reaction:
    15·1 answer
  • En la tabla periódica se encuentran diferentes tipos de elementos, entre ellos metales, no metales, metales de transición intern
    11·1 answer
  • Consider 55 mL of water (H2O) in a beaker and 55 mL of acetone [(CH3)2CO] in an identical beaker under identical conditions. Com
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!