What types of atoms typically form metallic bonds?

The combustion of 3.795 mg of liquid B, which contains only C, H, and O, with excess oxygen gave 9.708 mg of CO2 and 3.969 mg of

attashe74 [19]

Answer:

Explanation:

9.708 mg of CO₂ will contain 12 x 9.708 / 44 = 2.64 g of C .

3.969 mg of H₂O will contain 2 x 3.969 / 18 = .441 g of H .

mg of O in given liquid B = 3.795 - ( 2.64 + .441 ) = .714 mg

ratio of mg of C , H , O in the compound = 2.64 : .441 : .714

ratio of no of atoms of C , H , O in the compound

= 2.64 / 12 : .441 /1 : .714 / 16

= .22 : .44 : .0446

= .22 / .22 : .44 / .22 : .044 / .22

= 1 : 2 : .2

1 x 5 : 2 x 5 : .2 x 5

= 5 : 10 : 1

empirical formula of the compound = C₅H₁₀O

Volume of 89.8 mL at 1 .00 atm at 200⁰C

volume of gas at 1 atm and 0⁰C = 89.8 x 273 / 473 mL

= 51.83 mL

51.83 mL weighs .205 g

22400 mL will weigh .205 x 22400 / 51.83 g

= 88.6 g

So molecular weight = 88.6

Let molecular formula be (C₅H₁₀O)ₙ

molecular weight = n ( 5 x 12 + 10 + 16 )

= 86 n

86 n = 88.6

n = 1 approx

So molecular formula is same as empirical formula

C₅H₁₀O is molecular formula .

6 0

3 years ago

Balance the equation for this reaction in an acidic solution MnO-4 to NO-3+Mn+2

marin [14]

Answer

MnO₄ + 2H⁺ +3NO₂⁻ →3NO₃⁻ + Mn²⁺ +H₂O

Explanation

This is a redox reaction (oxidation-reduction reaction) which involves the transfer of electrons between two species. i.e

Mn + 6e⁻→Mn²⁺ (reduction)

3N³⁺- 6e⁻→3Mn⁵⁺(oxidation)

4 0

3 years ago

Read 2 more answers

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride ? atoms of boron.

maks197457 [2]

Answer:

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.

2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.

Explanation:

The molecular formula of boron trifluoride is $BF_3$ .

So, one mole of boron trifluoride has one mole of boron atoms.

1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.

The number of atoms in 2.20 moles of boron is:

One mole of boron has ---- $6.023x10^2^3$ atoms.

Then, 2.20 moles of boron has

- $=2.20 mol. x 6.023 x 10^2^3 atoms /1 mol\\=13.25x10^2^3 atoms$

2. Calculate the number of moles of BF3 in 5.35*1022 molecules.

$(5.35x10^2^2 molecules/6.023x10^2^3)x 1mol\\=0.0888mol$

One mole of boron trifluoride has three moles of fluorine atoms.

Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.

=0.266mol of fluorine atoms.

5 0

3 years ago

Based on the equation 3 Cu (s) + 8 HNO3 (aq) 3 Cu(NO3)2 (aq) + 2NO (g) + 4H2O (g) how many grams of Cu would be needed to react

eimsori [14]

Answer:

Mass = 381.28 g

Explanation:

Given data:

Number of moles of HNO₃ = 16 mol

Mass of Cu needed to react with 16 mol of HNO₃ = ?

Solution:

Chemical equation:

3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 4H₂O + 2NO

Now we will compare the moles of Cu with HNO₃ from balance chemical equation.

HNO₃ : Cu

8 : 3

16 : 3/8×16 = 6

Mass of Cu needed:

Mass = number of moles × molar mass

Mass = 6 mol × 63.546 g/mol

Mass = 381.28 g

4 0

3 years ago

The reaction between sodium and chlorine that forms table salt is shown

Olegator [25]

Answer: 2 Na (s) + Cl(g) -> 2 NaCl (s)

Explanation:

7 0

3 years ago