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Llana [10]
3 years ago
6

Which of the following is not an example of a molecule?

Chemistry
1 answer:
dybincka [34]3 years ago
4 0

Answer:

He

Explanation:

A molecule is the smallest physical unit of a substance that can exist independently, consisting of 2 or more atoms chemically combined

He is one atom of Helium

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Suppose 1.87g of nickel(II) bromide is dissolved in 200.mL of a 52.0mM aqueous solution of potassium carbonate.
Reika [66]

Answer:

0.0428 M

Explanation:

Because we're asked to calculate the molarity of nickel(II) cation, we need to <u>determine all sources for that species</u>, in this case, all Ni⁺² comes from the nickel(II) bromide solid (NiBr₂).

We use the molecular weight of NiBr₂ to calculate the moles of Ni:

1.87 g NiBr₂ ÷ 218.49g/mol * (1molNi⁺²/1molNiBr₂) =  8.55x10⁻³ mol Ni⁺²

Then we <u>divide the moles by the volume in order to calculate the concentration</u>:

8.55x10⁻³ mol Ni⁺² / 0.200 L = 0.0428 M

7 0
2 years ago
Calculate the molarity of the two solutions.
daser333 [38]

1. 0.33 M

2. 0.278 M

<h3>Further explanation</h3>

Molarity is a way to express the concentration of the solution

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

1. 0.350 mol of NaOH in 1.05 L of solution.

n=0.35

V=1.05 L

Molarity :

\tt M=\dfrac{0.35}{1.05}=0.33

2. 14.3 g of NaCl in 879 mL of solution.

mol NaCl(MW=58.5 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{14.3~g}{58.5~g/mol}=0.244

Molarity :

\tt M=\dfrac{0.244}{0.879~L}\\\\M=0.278

4 0
2 years ago
Aquatic organisms that lived hundreds of millions of years ago and were buried in silt and sediment resulted in the formation of
ololo11 [35]
The formation of Fossil Fuels.

The fossil fuels such as oil originate mostly from aquatic organism and are extracted on oil rigs located in oceans and seas. Most of the oil is procured this way.
3 0
2 years ago
During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final
kompoz [17]

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

7 0
2 years ago
Help me please I’ll give you brainest
Alika [10]

Answer:

I also want its answer

Please fast

4 0
2 years ago
Read 2 more answers
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