All categories

12 days ago

The atomic radius of magnesium is 150 pm. The atomic radius of strontium is 200 pm. What is the atomic radius of calcium

Chemistry

1 answer:

castortr0y [923]12 days ago

7 0

Answer:

Calcium's atomic radius is roughly 175 pm.

Explanation:

We know that magnesium has an atomic radius of 150 pm.

The atomic radius of strontium measures 200 pm.

Since calcium's position is between magnesium and strontium in group 2 of the periodic table, its atomic radius should be roughly averaged between magnesium's and strontium's atomic radii because atomic radius is not constant.

Thus;

Calcium's atomic radius is approximately calculated as follows;

The average atomic radius is (200 + 150)/2 = 175 pm.

You might be interested in

A chemist combined chloroform (CHCl3) and acetone (C3H6O) to create a solution where the mole fraction of chloroform is 0.187. T

KiRa [971]

Answer:

$\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}$

Explanation:

1. Molar concentration

Designate chloroform as C and acetone as A.

The molar concentration for C is derived from Moles of C per Litres of solution.

(a) Moles of C

We are assuming there are 0.187 moles of C.

This resolves that step.

(b) Litres of solution

Next, identify 0.813 moles of A.

(i) Mass of each component

$\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}$

(ii) Volume of each component

$\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}$

(iii) Volume of solution

Assuming mixing doesn't alter the total volume.

V = 15.08 mL + 59.70 mL = 74.78 mL

(c) Molar concentration of C

$c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}$

2. Molal concentration of C

Molal concentration is calculated as moles of solute per kilograms of solvent.

Total moles of C = 0.187 mol.

Mass of A = 47.22 g = 0.047 22 kg.

$\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}$

4 0

1 day ago

A crystallographer measures the horizontal spacing between molecules in a crystal. The spacing is

castortr0y [923]

Answer: The overall width of a crystal measures 1.65 mm.

Explanation:

Horizontal distance separating the two molecules is 16.5 nm.

Width of the $10^5$ molecules: $16.5 nm\times 10^5=16.5\times 10^5 nm$

$1 nm=10^{-6} mm$

The overall width of a crystal measured in millimeters= $16.5\times 10^5\times 10^{-6} mm=1.65 mm$

The overall width of a crystal is 1.65 mm.

7 0

11 days ago

Read 2 more answers

Consider butter (density= 0.860 g/mL) and sand (density= 2.28 g/mL). If 1.00 mL of butter were mixed with 1.00 mL of sand and mi

Anarel [852]

The mixture’s density is 1.57 g/cm³.

Step 1: Determine the mass of the butter.

$\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{0.680 g} }{\text{1 cm}^{3 }} = \text{0.860 g}\\$

Step 2: Determine the mass of the sand.

$\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{2.28 g} }{\text{1 cm}^{3 }} = \text{2.28 g}\\$

Step 3: Determine the density of the mixture.

Total mass = 0.860 g + 2.28 g = 3.14 g.

Total volume = 1 cm³ + 1 cm³ = 2 cm³

$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{3.14 g} }{\text{2 cm}^{3 }} = \textbf{1.57 g/cm}{^{3}\\$

6 0

1 day ago

Marianne designs an experiment involving electrically charged objects. She wants to know which objects will be attracted to a ne

castortr0y [923]

Answer:

The generation of static electricity occurs when two surfaces are rubbed together. This process causes a transfer of electrons, resulting in a build-up of negative charge. For instance, when you shuffle on a carpet, the friction creates multiple contact points which allow electrons to move onto you, thus accumulating a static charge. Touching another individual or object can lead to a sudden discharge, experienced as an electric shock.

In a similar way, rubbing a balloon against your hair generates opposite static charges on both your hair and the balloon. As you gently pull the balloon away from your head, the attraction between these opposite charges can be observed, causing your hair to rise.

Materials

• Balloon

• Woolen item (like a sweater, scarf, or yarn ball)

• Stopwatch

• Wall

• Partner (optional)

Preparation

• Inflate the balloon and secure the end.

• Have your partner ready to time with the stopwatch.

Procedure

• Grip the balloon with minimal hand coverage, such as holding it with just your thumb and index finger, or by its tied neck.

• Rub the balloon on the wool item once, making sure to go in one direction only.

• Press the rubbed side of the balloon against the wall and let go. Is it adhering to the wall? If it's stuck, your partner should start the stopwatch to measure how long it stays there. If it doesn’t stick, continue to the next step.

• Briefly touch the balloon to a metal object. Why is this step necessary?

• Repeat this procedure, but each time increase the number of rubs against the woolly item, ensuring the direction remains the same (do not rub back and forth).

Observations and results

As you increase the number of times you rub the balloon on the woolly material, does the duration of its adhesion to the wall increase?

Wool is an excellent conductor; it easily relinquishes electrons. When you rub wool on a balloon, electrons move from the wool to the surface of the balloon, imparting a negative charge to the rubbed area. Balloons, made from rubber, act as insulators, which means not all areas of the balloon will have a negative charge—only where it was rubbed will have a negative charge, while the rest of the balloon remains neutral.

Once the balloon is sufficiently charged negatively by repeated rubbing, it will adhere to the wall. Though the wall typically has a neutral charge, its internal charges can realign such that a positively charged region can attract the negatively charged balloon. Since the wall is also an insulator, the charge does not dissipate instantly. However, when the balloon is in contact with a metal object, the excess electrons from the balloon flow into the metal quickly, making the balloon lose its attraction and peel away.

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST

6 0

4 days ago

Read 2 more answers

One tank of gold fish is fed the normal amount of food once a day. A second tank is fed twice a day. A third tank is fed four ti

Tems11 [846]

The inquiry is incomplete; here is the full question:

One tank of goldfish receives the standard amount of feeding once daily, a second tank is given two feedings a day, and a third tank is fed four times daily throughout a six-week experiment. The body fat of the fish is recorded every day.

Independent Variable-

Dependent Variable-

Constants

Control Group-

Answer:

A) The quantity of food given to the goldfish

B) The body fat of the goldfish

C) -Type of fish in the experiment (goldfish)

Time period for feeding the fish (six weeks)

Shape and size of the tanks

D) group of goldfish receiving the standard feeding amount

Explanation:

The objective of the experiment is to assess how the quantity of food affects the body fat of goldfish. Consequently, the amount of food serves as the independent variable while the body fat acts as the dependent variable.

The control group is the one given the standard feeding amount (once daily). All subjects are goldfish, fed over a six-week duration, with all tanks being the same shape and size, establishing the constants in the research.

4 0

15 days ago