In experiments, variation is not present.<br> O True<br> O False

Calculate the number of molecules found in 35 g of Sodium Hydroxide?

Helen [10]

Answer:

5.27*10^23 (rounded to 3 significant figures)

Explanation:

The amount of molecules in one mole of anything is equal to Avogadro's number: 6.022×10^23

To find the number of moles of NaOH in 35 grams of it, do 35 divided by the molar mass (39.997): 35/39.997=0.87506562 moles of NaOH

To find the number of molecules, multiply the moles of NaOH by Avogadro's number: 0.87506562×(6.022×10^23)=5.26964522*10^23

7 0

3 years ago

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A 420 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 7 mL of 1.00 M KOH. What is the pH following this addition

Art [367]

<u>Answer:</u> The pH of the resulting solution will be 3.60

<u>Explanation:</u>

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$ ......(1)

We are given:

Molarity of formic acid = 0.100 M

Molarity of potassium formate = 0.100 M

Volume of solution = 420 mL = 0.420 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$\text{Moles of formic acid}=(0.100mol/L\times 0.420L)=0.0420mol$

$\text{Moles of potassium formate}=(0.100mol/L\times 0.420L)=0.042mol$

Molarity of KOH = 1.00 M

Volume of solution = 7 mL = 0.007 L

Putting values in equation 1, we get:

$\text{Moles of KOH}=(1mol/L\times 0.007L)=0.007mol$

The chemical equation for the reaction of formic acid and KOH follows:

$HCOOH+KOH\rightleftharpoons HCOOK+H_2O$

I: 0.042 0.007 0.042

C: -0.007 -0.007 +0.007

E: 0.035 - 0.049

Volume of solution = [420 + 7] = 427 mL = 0.427 L

To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:

$pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}}$ .......(2)

Given values:

$[HCOOK]=\frac{0.049}{0.427}$

$[HCOOH]=\frac{0.035}{0.427}$

$pK_a=3.75$

Putting values in equation 2, we get:

$pH=3.75-\log \frac{(0.049/0.427)}{(0.035/0.427)}\\\\pH=3.75-0.146\\\\pH=3.60$

Hence, the pH of the resulting solution will be 3.60

6 0

3 years ago

Ammonia, nh3, is a weak base with a kb = 1.8 x 10-5. in a 0.8m solution of ammonia, which has a higher concentration: ammonia (n

insens350 [35]

Answer is: concentration ammonia is higher than concentration of ammonium ion.
Chemical reaction of ammonia in water: NH₃ + H₂O → NH₄⁺ + OH⁻.
Kb(NH₃) = 1,8·10⁻⁵.
c₀(NH₃) = 0,8 mol/L.
c(NH₄⁺) = c(OH⁻) = x.
c(NH₃) = 0,8 mol/L - x.
Kb = c(NH₄⁺) · c(OH⁻) / c(NH₃).
0,000018 = x² / 0,8 mol/L - x.
solve quadratic equation: x = c(NH₄⁺) = 3,79·10⁻³ mol/L.

7 0

3 years ago

A solution of nitrous acid and potassium nitrite acts as a buffer due to reactions that occur within the solution when a strong

Ghella [55]

Answer:

a. NO₂⁻ + H⁺ → HNO₂

b. HNO₂ + OH⁻ → NO₂⁻ + H₂O

Explanation:

A buffer is defined as an aqueous mixture of a weak acid and its conjugate base or vice versa.

The buffer of the problem is HNO₂/NO₂⁻ <em>where nitrous acid is the weak acid and NO₂⁻ is its conjugate base.</em>

a. When a acid is added to a buffer as the buffer of the problem, the conjugate base will react with the acid, to produce the weak acid, thus:

NO₂⁻ + HCl → HNO₂ + Cl⁻

Ionic equation is:

NO₂⁻ + H⁺ + Cl⁻ → HNO₂ + Cl⁻

In the net ionic equation, you avoid the ions that don't react, that is:

b. In the same way, the weak acid will react with the strong acid producing water and the conjugate base, thus:

HNO₂ + NaOH → NO₂⁻ + H₂O + Na⁺

The ionic equation is:

HNO₂ + Na⁺ + OH⁻ → NO₂⁻ + H₂O + Na⁺

And the net ionic equation is:

5 0

3 years ago

Please help me ASAP!!! THANKS

Elan Coil [88]

Answer:

3

Explanation:

Nonmetals are only on the right side (other than hydrogen of course).

Malleable means to be able to be hammered out of shape with no cracking and luster is shiny (metallic properties)

Most likely a transition metal

6 0

3 years ago

In experiments, variation is not present. O True O False