Why is silicon dioxide a suitable material for lining furnaces

The "d" subshell can hold a maximum of ________ electrons.

Verizon [17]

The d subshell has 5 orbitals (with each being able to hold two electrons. As such, the d orbital can hold a maximum of 10 electrons.

The "d" subshell can hold a maximum of _TEN_ electrons.

5 0

2 years ago

The concentration of protein in a urine sample is calculated to be 2.77 μg/mL. What is the concentration of this solution in uni

maxonik [38]

Answer:

The concentration of this solution in units of pounds per gallon is $2.776*10^{-5} \frac{lb}{gal}$

Explanation:

Units of measurement are established models for measuring different quantities. The conversion of units is the transformation of a quantity, expressed in a certain unit of measure, into an equivalent one, which may or may not be of the same system of units.

In this case, the conversion of units is carried out knowing that 1 μg are equal to 2.205*10⁻⁹ Lb and 1 mL equals 0.00022 Gallons. So

$2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}$

If 1 μg equals 2.205*10⁻⁹ lb, 2.77 μg how many lb equals?

$lb=\frac{2.77ug*2.205*10^{-9}lb }{1ug}$

lb=6.10785*10⁻⁹

So, 2.77 μg= 6.10785*10⁻⁹ lb

Then:

$2.77 \frac{ug}{mL} = \frac{2.77 ug}{mL}=\frac{6.10785*10^{-9}lb }{mL} =\frac{6.10785*10^{-9}lb }{0.00022 gal} =\frac{6.10785*10^{-9}lb }{0.00022 gal}$

You get:

$2.77 \frac{ug}{mL} = 2.776*10^{-5} \frac{lb}{gal}$

The concentration of this solution in units of pounds per gallon is $2.776*10^{-5} \frac{lb}{gal}$

6 0

2 years ago

If 6 g of element k combine with 17 g of element l, how many grams of element k combine with 85 g of element l?

Ede4ka [16]

Hope this helps you.

5 0

3 years ago

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

25g of potassium nitrate are dissolved in 100g of water. What is the percent by mass of potassium nitrate?

trapecia [35]

The the exact mass is 24.740% (according to the internet) but I’d say that it’s 25

8 0

2 years ago