A sub-shell with n = 6, l = 2 can accommodate a maximum of:

Yasir wants to determine if there is a relationship between room color and sleep. Based on his research, Yasir makes an educated

Fynjy0 [20]

Yasir wanted to study the relationship between sleep and room color. So he has done all the background study required on the topic and arrived at a hypothesis that people fall asleep more rapidly in a room painted in blue color than in a room painted yellow. Yasir inquired several people which color they like better -yellow or blue- and used their feedback to decide whether his hypothesis was correct. There was no experiment conducted to observe the effect of a room color on sleep and the variables to be tested in the experiment were not properly identified.

Therefore, the correct answer would be,

An experiment that directly tests the hypothesis

Variables to be tested by an experiment

3 0

3 years ago

Read 2 more answers

Please help ill give you points and brainliest please

Paul [167]

Answer:

Answers with detail are given below

Explanation:

1) Given data:

Mass of Rb₃Rn = 76.19 g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 478.43 g/mol

Number of moles = 76.19 g/ 478.43 g/mol

Number of moles = 0.16 mol

2) Given data:

Mass of FrBi₂ = 120.02 g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 640.96 g/mol

Number of moles = 120.02 g/640.96 g/mol

Number of moles = 0.19 mol

3) Given data:

Mass of Zn₂F₃ = 88.24 g

Number of moles = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 187.73 g/mol

Number of moles = 88.24 g/ 187.73 g/mol

Number of moles = 0.47 mol

4) Given data:

Number of moles of Sb₄Cl = 1.20 mol

Mass of Sb₄Cl = ?

Solution:

Number of moles = mass/molar mass

Molar mass = 522.49 g/mol

Mass = Number of moles × molar mass

Mass = 1.20 mol × 522.49 g/mol

Mass = 626.99 g

3 0

3 years ago

What does lower case k stand for?

DaniilM [7]

Just for more clarification, lowercase k is the rate constant. Uppercase K is the equilibrium constant. You can actually use k to find K (equilibrium constant). K=k/k' This means that the equilibrium constant is the rate constant of the forward reaction divided by the rate constant of the reverse reaction

4 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

5.36 liters of nitrogen gas are at STP. What would be the new volume if we increased the moles from 3.5 moles to 6.0 moles?

Aleksandr [31]

Answer:

$V_2=9.20L$

Explanation:

Hello there!

In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

$\frac{V_2}{n_2}= \frac{V_1}{n_1}$

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

$V_2= \frac{V_1n_2}{n_1} \\\\V_2=\frac{5.36L*6.0mol}{3.5mol}\\\\V_2=9.20L$

Best regards!

3 0

3 years ago