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3 years ago

In the chemical equation 3c2h4 how many atoms of carbon are represented

1 answer:

5 0

An example.
water is H2O

2 hydrogen, 1 oxygen

so the number to the right means how much of what is on the left.

so it looks like 2, because C2, but look at the 3 at the beginning. that means
3 (c2h4)

so 6 carbons, 12 hydrogen

the ratio of c2 to h4 doesn't change it's always 1:2.

but the 3 at the front is a different number relating to how much you have

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Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration

elixir [45]

Answer:

8.54

Explanation:

At equivalence point :

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052 = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :

pH = 1/2 [ pKw + pKa + log c ]

Ka for acetic acid = 1.75 X 10⁻⁵

so pKa = -log (1.75 X 10⁻⁵) = 4.74

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169

so log c = log 0.02169 = -1.66

putting the values....

pH = 1/2 [14 + 4.74 - 1.66 ]

pH = 1/2 [ 17.08] = 8.54

6 0

3 years ago

Read 2 more answers

A sample of n2 effuses in 255 s. how long will the same size sample of cl2 take to effuse?

7nadin3 [17]

For this problem, we use Graham's Effusion Law to find out the rate of effusion of chlorine gas. The formula is as follows:

R₁/R₂ = √(M₂/M₁)

Let 1 be N₂ while 2 be Cl₂

255/R₂ = √(28/70.8)
Solving for R₂,
R₂ = 405.5 s

<em>Thus, it would take 405.5 s to effuse chlorine gas.</em>

4 0

3 years ago

Describe the main differences beA human cell has 46 chromosomes. At the end of mitosis, there are two cells, each with 46 chromo

liberstina [14]

Answer:

At the end of meiosis, there are four cells, each with 23 chromosomes, for a total of 92 chromosomes split between the four cells.

Explanation:

During meiosis, a diploid cell (46 chromosomes) replicates its DNA (making 92 chromosomes) then undergoes two cell divisions to generate four haploid cells (23 chromosomes).

These haploid cells are the gametes which, during fertilization, fuse to become a zygote with 46 chromosomes.

6 0

3 years ago

A car travels 10km in 10 min so what is the average

ipn [44]

Average speed in km/h:

S = D/T

T = 10/60 hrs
= 0.1667

S = 10 / 0.1667
= 59.988 km/h

4 0

3 years ago

The teacher tells your group to make a stock solution of sodium chloride, and then diluting it to

sergejj [24]

The idea here is that you need to figure out how many moles of magnesium chloride,

MgCl

, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.

As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.

So, how many moles of magnesium chloride must be present in the target solution?

⇒

⋅

0.158 M

⋅

250.0

⋅

−

0.0395 moles MgCl

Now determine what volume of the target solution would contain this many moles of magnesium chloride

⇒

0.0395

moles

3.15

moles

0.01254 L

Rounded to three sig figs and expressed in mililiters, the volume will be

12.5 mL

So, to prepare your target solution, use a

12.5-mL

sample of the stock solution and add enough water to make the volume of the total solution equal to

250.0 mL

This is equivalent to diluting the

12.5-mL

sample of the stock solution by a dilution factor of

3 0

2 years ago