How many grams of CO2 could be formed with 2.09 x 1023 atoms of O?

What must occur in order for a chemical reaction to be specifically classified as endothermic instead of exothermic?

Alla [95]

Answer: B- Chemical bonds are formed. Energy is released in the form of heat.

Explanation: I hoped that helped !

6 0

3 years ago

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What is common to all elements in the same group on the periodic table? Their properties are very different.

Mrac [35]

Hello,

B.They have the same number of valence electrons.

7 0

3 years ago

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In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o

Zigmanuir [339]

Answer:

Stock solution = 1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

Step 1: Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

Step 2: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

Step 3: Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

Step 4: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

Step 5: Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

Step 6: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

5 0

3 years ago

What is the percent composition HCN?

kupik [55]

Answer:

see calculations in explanation

Explanation:

percent = part/total x 100%

part = ∑ atomic mass of element

hydrogen = 1.008 amu (atomic mass units)
carbon = 12.011 amu
nitrogen = 14.007 amu

total = ∑ molecular mass of compound

= H amu + C amu + Namu

= 1.008 amu + 12.011 amu + 14.007 amu

= 27.026 amu

%H = (1.008amu/27.026amu)100% = 3.730%

%C = (12.011amu/27.026amu)100% = 44.442%

%N = (14.007amu/27.026amu)100% = 51.827%

Check results ∑%values = 100%

3.730% + 44.442% + 51.827% = 99.999% ≅ 100%

7 0

3 years ago

If one adds 0.1 mol of the weak acid hf (pk_a = 3.2) to a solution with a ph = 2, which species would be most abundant and how m

zavuch27 [327]

$F^{-}$ is most abundant and 6310 times more than HF.

<h3>What is a strong and weak acid?</h3>

When an acid is dissolved in water, all of its molecules disintegrate, making the acid powerful.

When an acid is dissolved in water, only a small number of its molecules disintegrate, making the acid weak. Strong acids have a lower pH than weak acids.

The powerful acids include perchloric acid, chloric acid, nitric acid, sulfuric acid, hydrobromic acid, and hydroiodic acid.

Given:

Pka=3..2

pH=7

Let the volume be 1 liter

[HF]=01 M

$pH=pka+log \frac{F^{-}}{HF} \\\\7=3.2+log\frac{F^{-}}{HF} \\3.8=log\frac{F^{-}}{HF}\\ \frac{F^{-}}{0.1}=10^{3.8} \\F^{-}=630.95 M$

Now,

$\frac{F^{-}}{HF}=\frac{630.95}{0.1}\\ =6309.57$

F-:HF= 6309.57:1

Therefore, the most abundant is $F^{-}$ and has 6310 times more than HF is $F^{-}$ .

To know more about strong and weak acids, visit: brainly.com/question/12811944

#SPJ4

4 0

1 year ago