Based on the Lewis/electron dot representation

How many moles of water vapor are formed. when 10 litres of butane gas c4h10 is burned in oxygen at STP

NeX [460]

Answer:

gto

Explanation:

4 0

3 years ago

You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2

avanturin [10]

1) Answer is: the formula is C₇N₃O₆H₅.

M(TNT) = 0.150 g; mass of the trinitrotoluene.

ω(N) = 18.5% ÷ 100%.

ω(N) = 0.185; mass percentage of the nitrogen.

m(N) = 0.150 g · 0.185.

m(N) = 0.02775 ·; mass of the nitrogen.

n(N) = 0.02775 g ÷ 14 g/mol.

n(N) = 0.002 mol; amount of the nitrogen.

n(CO₂) = 0.204 g ÷ 44 g/mol.

n(CO₂) = 0.0046 mol.

n(C) = n(CO₂) = 0.0046 mol; amount of the carbon.

m(C) = 0.0046 mol · 12 g/mol.

m(C) = 0.0552 g; mass of the carbon.

n(H₂O) = 0.030 g ÷ 18 g/mol.

n(H₂O) = 0.00166 mol.

n(H) = 2 · n(H₂O) = 0.0033 mol; amount of the hydrogen.

m(H) = 0.0033 mol · 1 g/mol.

m(H) = 0.0033 g; mass of the hydrogen.

2) m(O) = m(TNT) - m(N) - m(C) - m(H).

m(O) = 0.150 g - 0.02775 g - 0.0552 g - 0.0033 g.

m(O) = 0.06375 g.

n(O) = 0.06375 g ÷ 16 g/mol.

n(O) = 0.004 mol; amount of oxygen.

n(C) : n(N) : n(O) : n(H) = 0.0046 mol : 0.002 mol : 0.004 mol : 0.0033 mol.

n(C) : n(N) : n(O) : n(H) = 2.33 : 1 : 2 : 1.66 /×3.

n(C) : n(N) : n(O) : n(H) = 7 : 3 : 6 : 5.

8 0

3 years ago

Boron trifluoride gas is collected at 2.0 degree C in an evacuated flask with a measured volume of 15.0 L. When all the gas has

IgorC [24]

Answer:

Moles of boron trifluoride gas that were collected = 11.6 mol

Mass of boron trifluoride gas that were collected = 787 g

Explanation:

Given that:

Temperature = 2.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (2.0 + 273.15) K = 275.15 K

V = 15.0 L

Pressure = 0.130 atm

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

0.130 atm × 15.0 L = n ×0.0821 L atm/ K mol × 275.15 K

⇒n = 11.6 mol

Thus, Moles of boron trifluoride gas that were collected = 11.6 mol

Molar mass of boron trifluoride gas = 67.82 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$11.6\ mole= \frac{Mass}{67.82\ g/mol}$

Mass of boron trifluoride gas that were collected = 787 g

7 0

3 years ago

The Ksp for Cu(OH)2 is 4.8 × 10-20. Determine the molar solubility of Cu(OH)2 in a buffer solution with a pH of 10.1.

Papessa [141]

Answer:

E) 3.0x10⁻¹²

Explanation:

Ksp of Cu(OH)₂ is defined as:

Cu(OH)₂(s) ⇄ Cu²⁺(aq) + 2 OH⁻(aq)

Ksp = [Cu²⁺] [OH⁻]²

When pH is 10.1, [OH⁻] is:

pOH = 14 - pH

pOH = 3.9

pOH = -log [OH⁻]

[OH⁻] = 1.26x10⁻⁴M

Replacing in ksp formula:

4.8x10⁻²⁰= [Cu²⁺] [1.26x10⁻⁴]²

3.0x10⁻¹² = [Cu²⁺]

That means the maximum amount of Cu²⁺ that can be in solution is 3.0x10⁻¹²M, thus, molar solubility of Cu(OH)₂ is

E) 3.0x10⁻¹²

3 0

3 years ago

Arange the following elements, from top to bottom, in the correct order of increasing atomic radius starting with the smallest.

Sever21 [200]

Answer:

S 16

Rh 45

In 49

Sb 51

Te 52

Explanation:

I wasn't sure what you meant by St, but i think you meant Sb. Let me know if I'm mistaken

7 0

3 years ago

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