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Issue like gay marriage, abortion, and education have caused debate over the constitution because of the issue of what? (Hint: f

Korolek [52]

Yes 81671 cupcake ham shops

3 0

2 years ago

What is the noble-gas configuration for As

jasenka [17]

Answer:

Noble gas Electronic configuration of arsenic:

As₃₃ = [Ar] 3d¹⁰ 4s² 4p³

Explanation:

Arsenic is metalloid.

Its atomic number is 33.

Its atomic mass is 75 amu.

Its symbol is As.

It is usually present in combine with sulfur and metals.

it is used in bronzing.

It is also used for hardening.

Electronic configuration:

As₃₃ = Is² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³

Noble gas Electronic configuration:

As₃₃ = [Ar] 3d¹⁰ 4s² 4p³

Noble gas electronic configuration is shortest electronic configuration by using the noble gas elements full octet electronic configuration.

3 0

3 years ago

You decide to clean the bathroom. You notice that the shower is covered in a strange green slime . you try to get rid of this sl

Free_Kalibri [48]

The independent variable (IV) is the lemon juice mixture
The dependent variable (DV) is the appearance of the green slime on the shower
The control variable (CV) are time taken to spray, the amount of spray
The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
The control group (CG) is the side of the shower sprayed with water.

INDEPENDENT VARIABLE

Independent variable is the variable of an experiment that is changed by the experimenter in order to bring about a change. It is the variable being tested in the experiment. In this case, the IV is the lemon juice mixture tested on the green slime on the shower.

DEPENDENT VARIABLE:

Dependent variable is the variable that is observed or measured in an experiment. It is also called responding variable. The DV in this case is the appearance of the green slime on the shower.

CONTROL VARIABLE:

Control variable is the variable that is kept constant throughout the experiment for all groups. The CV is the same for all the groups and they include: time taken to spray, the same amount of spray

CONTROL GROUP

Control group is the group that does not receive the independent variable or test in an experiment. In this case, the CG is the side of the shower sprayed with water.

EXPERIMENTAL GROUP:

Experimental group is the group of ab experiment that receives the experimental treatment or independent variable. In this case, the EG is the side of the shower sprayed with lemon juice mixture.

Therefore, the IV, DV, CV, EG and CG of this experiment are as follows:

The independent variable (IV) is the lemon juice mixture
The dependent variable (DV) is the appearance of the green slime on the shower
The control variable (CV) are time taken to spray, the amount of spray
The experimental group (EG) is the side of the shower sprayed with lemon juice mixture
The control group (CG) is the side of the shower sprayed with water.

Learn more: brainly.com/question/17498238?referrer=searchResults

7 0

3 years ago

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

5 0

3 years ago

Question 7 (1 point)

Anastasy [175]

Answer:

2.28bar

Explanation: Boyle's law P1V1=P2V2 manipulate formula in favor of V2 the new formula should beP1V1/V2

0.895*318/125=2.2768 but to same sigfig it is 2.28

5 0

3 years ago