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3 years ago

Using the periodic table determine the atomic mass of Na2C2O4

1 answer:

4 0

Masses of atoms are the sum of neutrons and protons. Atomic mass given for the element is the weighted atomic masses of the isotopes depending on the abundance of the isotopes.
Atomic masses of the elements making up Na₂C₂O₄ are as follows;
Na - 22.98 a.m.u
2 atoms of Na - 22.98 x 2 = 45.96
C - 12.01 a.m.u
2 atoms of C - 12.01 x 2 = 24.02
O - 15.99 a.m.u
4 atoms of O - 15.99 x 4 = 63.96
Sum of the atomic masses = 45.96 + 24.02 + 63.96 = 133.94
Mass is 133.94 g/mol

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

If the formula mass of one molecule is X amu, the molar mass is X g/mol. Group of answer choices True False

Irina-Kira [14]

yes

Explanation:

the molar mass of a compound is g/mol

8 0

1 year ago

How many grams of water can I convert from a liquid to a gas with 6768 joules?

vredina [299]

Answer:

The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g

Explanation:

The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid

The latent heat of evaporation of water, Δ $H_v$ ≈ 2,230 J/g

The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows

Q = m × Δ $H_v$

Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;

6,768 J = m × 2,230 J/g

∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g

The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.

8 0

3 years ago

A student determines that a 9.8 g mixture of MgCl₂ (s) and NaNO3(s) contains 0.050 mol of

Damm [24]

mass MgCl₂ = mol x MM MgCl₂ = 0.05 x 95.211 g/mol = 4.76 g

mass Cl in MgCl₂ :

= (2 x AM Cl)/MM MgCl₂ x mass MgCl₂

= (2 x 35.5 g/mol)/95.211 g/mol x 4.76

= 3.55 g

% mass Cl in the mixture :

= (mass Cl / mass mixture) x 100%

= 3.55 / 9.8 x 100%

= 36.22%

6 0

1 year ago

The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?

Vitek1552 [10]

The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
[OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
NaA → Na+ + A-
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
A- + H2O ⇌ HA + OH-
Initial 0.0100 0 0
Change -x +x +x
Equilibrium 0.0100-x x x

We can now calculate the Kb for A-:
Kb = [HA][OH-] / [A-]
= x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
Ka = Kw / Kb
= 1.00x10^-14 / 1.00x10^-4
= 1.00x10^-10

7 0

3 years ago