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Juliette [100K]
3 years ago
9

PLZ HELLLLPPPP

Chemistry
1 answer:
coldgirl [10]3 years ago
6 0

D. The nucleus is the control center of the cell


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How many of the planets have an orbital period of less than one Earth year?
GrogVix [38]

Since orbital period depends on how far you are from the sun, planets closer to the sun have a orbital period less than one earth year.

These planets are Mercury and Venus

8 0
3 years ago
For all of the following questions 20.00 mL of 0.200 M HBr is titrated with 0.200 M KOH.
HACTEHA [7]

Answer :

The concentration of H^+ before any titrant added to our starting material is 0.200 M.

The pH based on this H^+ ion concentration is 0.698

Explanation :

First we have to calculate the concentration of H^+ before any titrant is added to our starting material.

As we are given:

Concentration of HBr = 0.200 M

As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion H^+ and bromide ion Br^-.

As, 1 M of HBr dissociates to give 1 M of H^+

So, 0.200 M of HBr dissociates to give 0.200 M of H^+

Thus, the concentration of H^+ before any titrant added to our starting material is 0.200 M.

Now we have to calculate the pH based on this H^+ ion concentration.

pH : It is defined as the negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

pH=-\log (0.200)

pH=0.698

Thus, the pH based on this H^+ ion concentration is 0.698

3 0
3 years ago
Rocks are solids composed of one or more minerals true or false
Yuki888 [10]
The answer is true
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8 0
3 years ago
4. The type of thermal energy transfer that heats our homes is called
andrew-mc [135]

Answer:

conduction is the answer

6 0
3 years ago
You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli
maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

M_1V_1=M_2V_2

where, M_1 is initial molarity and  M_2 is the molarity after dilution. Similarly,  V_1 is the volume before dilution and  V_2 is the volume after dilution.

Let's plug in the values in the equation:

1.25M(363mL)=0.50M(V_2)

V_2=\frac{1.25M(363mL)}{0.50M}

V_2=907.5mL

Volume of water added = 907.5mL - 363mL  = 544.5 mL

So, 544.5 mL of water are need to be added to the original solution for dilution.

3 0
3 years ago
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