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AURORKA
13 days ago
11

88 g of the strontium reacts with 160 g bromine to produce how many strontium bromide

Chemistry
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The table compares some characteristics of two atoms.
alisha [2963]
Atoms X and Y are placed in identical positions in the periodic table because they are isotopes.
7 0
2 months ago
Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
lorasvet [2795]

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles

Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

4 0
4 months ago
Read 2 more answers
One cubic millimeter (mm3) of blood contains 7.0 x 106 red blood cells. How many red blood cells are in 1.0 L of blood?
Anarel [2989]
The problem provides a conversion factor---> 1 mm3= 7.0 x 10^6 RBC. Therefore, to determine the quantity of red blood cells in your sample, we must first convert Liters to cm3 using the conversion factor--> 1 mL= 1 cm3

I have shared how to resolve this issue.


7 0
2 months ago
(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o
Tems11 [2777]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100

Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100

Percent Ionic character = 1 - e^(^-^0^.^1^) * 100

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

8 0
4 months ago
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