Elements in the 2nd family/group gain or lose electrons? *

What natural and human made features can a map show? Give two examples of each.

Bess [88]

Natural: Land mounds, Rivers
Human made: Country borders, cities

3 0

3 years ago

How many grams are in 4.07x10^15 molecules of calcium hydroxide

maksim [4K]

There are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the number of moles in the substance by its molecular mass.

However, given that the number of molecules in calcium hydroxide is 4.07 x 10¹⁵molecules, we need to calculate the number of moles in Ca(OH)2 as follows:

no. of moles of Ca(OH)2 = 4.07 x 10¹⁵ ÷ 6.02 × 10²³

no. of moles = 0.676 × 10-⁸

no. of moles = 6.76 × 10-⁹ moles.

Molar mass of Ca(OH)2 = 74.093 g/mol

Mass of Ca(OH)2 = 74.093 × 6.76 × 10-⁹ moles

Mass = 5.01 × 10-⁷grams.

Therefore, there are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.

Learn more about how to calculate mass at: brainly.com/question/8101390?referrer=searchResults

5 0

2 years ago

Can someone please help me and ill mark you as brainlest

r-ruslan [8.4K]

Answer:

can you show the question so we can read it

3 0

3 years ago

Read 2 more answers

Zinc chloride + Magnesium →

Gelneren [198K]

Answer:

im assuming you need the balanced equation

Explanation:

Zn + MgCl2 = ZnCl2 + Mg

3 0

3 years ago

Read 2 more answers

Suppose 11.4g of ammonium chloride is dissolved in 250 ml of a 0.3M aqueous solution of potassium carbonate. Calculate the final

Sindrei [870]

Answer:

The molarity of the final ammonium cation is 0.252M

Explanation:

Step 1: Data given

Mass of ammonium chloride (NH4Cl) = 11.4 grams

Volume of 0.3 M aqueous solution of potassium carbonate (K2CO3) = 250 mL = 0.250L

Step 2: The balanced equation

2NH4Cl + K2CO3 → 2KCl + (NH4)2CO3

Step 3: Calculate moles of (NH4)Cl

moles (NH4)Cl = 11.4 grams /53.49 g/mol

Moles (NH4)Cl = 0.213 moles

Step 4: Calculate moles of K2CO3

Moles K2CO3 = Molarity * Volume

Moles K2CO3 = 0.3M * 0.250 L = 0.075 moles

Step 5: Calculate moles (NH4)Cl at the equilibrium

For 2 moles (NH4)Cl consumed, we need 1 mole of K2CO3 to produce 2 KCl and 1 mole of (NH4)2CO3

(NH4)2CO3l will dissolve in 2NH4+ + CO32-

Moles (NH4)2Cl = 0.213 moles - 2*0.075 = 0.063 moles

Moles NH4+ = moles (NH4)Cl = 0.063 moles

Step 6: Calculate Molarity of NH4+

Molarity = Moles / volume

Molarity of NH4+ = 0.063 moles / 0.250 L

Molarity of NH4+ = 0.252 M

The molarity of the final ammonium cation is 0.252M

5 0

3 years ago