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2 years ago

Elements that can exist in two or more geometric patterns are called

Chemistry

1 answer:

lesya [120]2 years ago

4 0

Answer:

The answer is allotropes

Explanation:

As known by studying and research the ansqer came out to be allotropes

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Which of the following reactions will not occur as written? Which of the following reactions will not occur as written? Sn (s) +

galben [10]

Answer:I have no idea

Explanation:

Why is this so complex…

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2 years ago

Read 2 more answers

Vinegar is an aqueous solution of acetic acid, ch3cooh. a 5.00 ml sample of a particular vinegar requires 26.90 ml of 0.175 m na

Fed [463]

The molarity of acetic acid in the vinegar is 0.94 M

<u><em> calculation</em></u>

Step 1: write the balanced equation between CH3COOH + NaOH

that is CH3COOH + NaOH → CH3COONa + H2O

step 2 : find the moles of NaOH

moles =molarity x volume in L

volume in liters = 26.90/1000=0.0269 l

moles = 0.175 mol /L x 0.0269 L =0.0047 moles of NaOH

Step 3: use the mole ratio to find moles of CH3COOH

that is the mole ratio of CH3COOH: NaOH is 1:1 therefore the moles of CH3COOH is =0.0047 moles

Step 4: find the molarity of CH3COOH

molarity = moles/volume in liters

volume in liter = 5.00/1000 =0.005 l

molarity is therefore=0.0047 moles/ 0.005 l = 0.94 M

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2 years ago

Circle the molecule(s) in the atmosphere that come from animals. (circle all that apply)

bagirrra123 [75]

Answer:

A and C

Explanation:

A; Animals inhale and exhale, breath oxygen, leave carbon dioxide.

C; We fart, so do they! (farts are a release of methane gas from our bodies) Just ask my dog...

Hope this helped!

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2 years ago

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine, CH3NH2, with 20.00 mL of 0.10 M methylammonium c

olya-2409 [2.1K]

Answer:

pH=10.97

Explanation:

the solution of methyl amine with methylammonium chloride will make a buffer solution.

The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:

$pOH=pKb+log\frac{[salt]}{[base]}$

pH = 14- pOH

Let us calculate pOH $pOH=3.43+ (-0.397)=3.03$

$pH=14-pOH=14-3.03=10.97$

[Salt] = [methylammonium chloride] = 0.10 M (initial)

After adding base

$[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M$

[base] = [Methylamine]=0.10

After mixing with salt

$[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M$

pKb= -log[Kb]= 3.43

Putting values

pOH = $3.43+log(\frac{[0.0286]}{0.0714}$

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2 years ago

A 250. ml sample of 0.0328m hcl is partially neutralized by the addition of 100. ml of 0.0245m naoh. find the concentration of h

Ksenya-84 [330]

Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution : 0.0328 M

$Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0328=\frac{\text{Number of moles}}{0.250 L}$

Moles of HCl in 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution : 0.0245 M

$Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0245=\frac{\text{Number of moles}}{0.100 L}$

Moles of NaOH in 0.100 L solution = 0.00245 moles

3) Concentration of hydrochloric acid in the resulting solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.

Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L

Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles = 0.00575 moles

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in L}}$

Molarity of HCl left un-neutralized : $\frac{0.00575 moles}{0.350L}=0.0164 M$

0.0164 molar concentration of hydrochloric acid in the resulting solution.

3 0

2 years ago