Ms. Sharkey wanted to see which tennis shoes made her run the fastest. She bought

How do you know if a reaction is exothermic

melisa1 [442]

When there is heat released

7 0

3 years ago

Read 2 more answers

What is Polymerase Chain Reaction?

KatRina [158]

Answer:

Polymerase Chain reaction is a method used by scientists to making A LOT (millions to billions) of copies of a DNA sample which then allows scientists to take a small of that DNA and increase the amount to a very large amount to study the DNA into further detail.

8 0

4 years ago

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A mixture consisting of only iron(II) chloride (FeCl2) and chromium(II) chloride (CrCl2) weighs 1.0294 g. When the mixture is di

LenKa [72]

Answer:

Mass FeCl2 = 0.0333g

Mass CrCl2 = 0.9961g

Explanation:

To solve this problem. The first equation we can write is:

Mass FeCl2 + Mass CrCl2 = 1.0294g (1)

Now, the Chlorides of FeCl2 and CrCl2 react producing 2.3609g of AgCl

Using molar mass of these species (126.75g/mol, 122.9g/mol, 143.32g/mol, respectively), you can write the equation:

2Mass FeCl2 / 126.75 + 2Mass CrCl2 / 122.9 = 2.3609/143.32

That is: Moles Chloride before = Moles Chloride in AgCl after reaction

7.8895x10⁻³Mass FeCl2 + 0.0162734MassCrCl2 = 0.016473 (2)

Replacing (1) in (2):

7.8895x10⁻³ (1.0294g - MassCrCl2) + 0.0162734MassCrCl2 = 0.016473

8.12145x10⁻³ -7.8895x10⁻³MassCrCl2 + 0.0162734Mass CrCl2 = 0.016473

8.3839x10⁻³ MassCrCl2 = 8.35155x10⁻³

Mass CrCl2 = 0.9961g

And:

Mass FeCl2 = 1.0294g - 0.9961g

Mass FeCl2 = 0.0333g

3 0

3 years ago

A 35.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following v

marissa [1.9K]

Answer:

a) 2.79

b) 4.75

c) 6.59

d) 8.81

e) 11.03

f) 12.42

Explanation:

Step 1: Data given

Volume of acetic acid = 35.0 mL

Molarity of acetic acid = 0.150 M

Molarity of NaOH = 0.150 M

Step 2: pH after adding 0 mL of NaOH

HA + H2O ⇆ A- + H3O+

Initial concentration

[HA] = 0.150 M

[A-] = 0M

[H3O+] = 0M

Concentration at the equilibrium

[HA] = 0.150 - x M

[A-] = xM

[H3O+] = xM

Ka = [H3O+][A-] / [HA] = x²/(0.150 -x) = 1.76*10^-5

0.150 >>> x

We can write it like this

x²/(0.150) = 1.76*10^-5

x² =0.00000264

x = 0.00162

[H3O+] = 0.00162 M

pH = -log [H3O+] = -log (0.00162) = 2.79

Step 3: pH after adding 17.5 mL of NaOH

This is the point of half-neutralization.

pH = pKa = -log (1.76*10^-5- 4.75

Step 4: pH after adding 34.5 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.0345 L

Moles NaOH = 0.005175 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.005175 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.005175 = 0.000075 moles

[OH-] = 0.005175 - 0.005175 = 0 moles

[A-] = 0.005175 moles

pH = pKa + log([A-]/[HA]

pH = 4.75 +log(0.005175/0.000075)

pH = 4.75 + 1.84 = 6.59

Step 4: pH after adding 35.0 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.035 L

Moles NaOH = 0.00525 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.00525 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.00525 = 0 moles

[OH-] = 0.00525 - 0.00525 = 0 moles

[A-] = 0.00525 moles

We have a solution of 0.00525 mol of A - in 70.0 mL

[A-] = 0.00525 moles / 0.070 L = 0.075 M

A- + H20 ⇆ HA + OH-

Initial moles A- = 0.0750 moles

initial moles HA = 0 moles

Initial moles OH- = 0 moles

Moles A- at the equilibrium = 0.0750 - x

Moles HA at the equilibrium = x

Moles OH- at the equilibrium = x

Kb = Kw/Ka = 10^-14 / 1.76*10^-5

Kb = 5.68 *10^-10

Kb = [HA][OH-] / [A-] = x² / 0.0750 - x = 5.68*10^-10

Since 0.0750 >>> x we can write

5.68*10^-10 = x²/0.0750

x = 6.53 *10^-6

[OH-] = x = 6.53 * 10^-6 M

pOH = - log (6.53*10^-6) = 5.19

pH = 14 - pOH = 14 - 5.19 = 8.81

Step 5: pH after adding 35.5 mL of NaOH

We add an excess moles of NaOH

We have an excess of 0.5 mL NaOH

Moles excess = 0.0005 L *0.150 M = 0.000075 moles

[OH-] = 0.000075 moles / (0.035+0.0355L)

[OH-] = 1.06 * 10^-3 M

pOH = -log( 1.06 *10^-3)

pOH = 2.97

pH = 14 -2.97 = 11.03

Step 6: pH after adding 50.0 mL of NaOH

We have an excess of 15.0 mL NaOH

Moles excess = 0.015 L *0.150 M = 0.00225 moles

[OH-] = 0.00225 moles / (0.035+0.050L)

[OH-] = 0.0265 M

pOH = -log( 0.0265M)

pOH = 1.58

pH = 14 -1.58 = 12.42

4 0

3 years ago

An element X has the proton number 18.

storchak [24]

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the element X with the atomic number of 18 is Argon (Ar) while the next element (element Y) in the periodic table, with atomic number 19, is potassium (K).

Element X is found in the group zero (also called group 18 and elements found therewith are referred to as noble gases) of the periodic table while element Y is found in group 1 (elements found here are referred to as alkali metals).

Element X has a completely filled outermost shell while element Y has one electron in its outermost shell.

Element X is a metal while element Y is a nonmetal.

8 0

3 years ago