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3 years ago

Determine the volume of methane (ch 4 ) gas needed to react completely with 0.660 l of o 2 gas to form methanol (ch 3 oh).

1 answer:

4 0

Balance Chemical Equation for this reaction is,

2 CH₄ + O₂ → 2CH₃OH

According to this eq, 22.4 L (1 moles) of Oxygen requires 44.8 L (2 mole) CH₄ for complete reaction.
So, the volume of CH₄ required to consume 0.66 L of O₂ is calculated as,

22.4 L O₂ required to consume = 44.8 L CH₄
0.660 L O₂ will require = X L of CH₄

Solving for X,
X = (44.8 L × 0.660 L) ÷ 22.4 L

X = 1.320 L of CH₄

Result:
1.320 L of CH₄ gas is needed to react completely with 0.660 L of O₂ gas to form methanol (CH₃OH).

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In Rutherfords gold foil experiment, what caused the deflection of the positively charged particles?

dmitriy555 [2]

The particles that completely went through passed through the empty space in the atom. The particlesthat were deflected to the sides were deflated by a passing electron that moved the alpha particle out of the way. The particles that were bounced back hit the nucleus in the center that it bounced off.


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3 0

3 years ago

Read 2 more answers

Initially, [NH3(g)] = [O2(g)] = 3.60 M; at equilibrium [N2O4(g)] = 0.60 M. Calculate the equilibrium concentration for NH3.

natima [27]

The question is incomplete, here is the complete question:

Consider the reaction $4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)$

Initially, $[NH_3(g)]=[O_2(g)]$ = 3.60 M; at equilibrium $[N_2O_4(g)]=0.60M$ . Calculate the equilibrium concentration for $NH_3$

Answer: The equilibrium concentration of ammonia is 2.8 M

Explanation:

We are given:

Initial concentration of $[NH_3(g)]$ = 3.60 M

Initial concentration of $[O_2(g)]$ = 3.60 M

For the given chemical equation:

$4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)$

Initial: 3.60 3.60

At eqllm: 3.60-4x 3.60-7x 2x 6x

We are given:

Equilibrium concentration of $[N_2O_4(g)]$ = 0.60 M

Evaluating the value of 'x'

$\Rightarrow 2x=0.60\\\\\Rightarrow x=\frac{0.60}{2}=0.2$

So, equilibrium concentration of $NH_3=(3.60-4x)=(3.60-(4\times 0.2))=2.8M$

Hence, the equilibrium concentration of ammonia is 2.8 M

6 0

3 years ago

Consider this equation: 2.524 g (5.1 × 106 g) ÷ (6.85 × 103 g) = ? How many significant figures should the result have?

weqwewe [10]

Answer:

Explanation:

2.524g(5.1)(106)g

(6.85)(103)g

1364.4744g2

705.55g

1364.4744g

705.55

1364.4744g

705.55

=1.933916g

6 0

4 years ago

The Mond process produces pure nickel metal via the thermal decomposition of nickel tetracarbonyl: Ni(CO)4 (l) → Ni (s) + 4CO (g

Yuki888 [10]

Answer: The volume of CO formed is 254.43 L.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of $Ni(CO)_4$ = 444 g

Molar mass of $Ni(CO)_4$ = 170.73 g/mol

Putting values in above equation, we get:

$\text{Moles of }Ni(CO)_4=\frac{444g}{170.73g/mol}=2.60mol$

For the given chemical reaction:

$Ni(CO)_4(l)\rightarrow Ni(s)+4CO(g)$

By stoichiometry of the reaction:

1 mole of nickel tetracarbonyl produces 4 moles of carbon monoxide.

So, 2.60 moles of nickel tetracarbonyl will produce = $\frac{4}{1}\times 2.60=10.4mol$ of carbon monoxide.

Now, to calculate the volume of the gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of the gas = 752 torr

V = Volume of the gas = ? L

n = Number of moles of gas = 10.4 mol

R = Gas constant = $62.364\text{ L Torr }mol^{-1}K^{-1}$

T = Temperature of the gas = $22^oC=(273+22)K=295K$

Putting values in above equation, we get:

$752torr\times V=10.4mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 295K\\\\V=254.43L$

Hence, the volume of CO formed is 254.43 L.

5 0

4 years ago

Amanda pushes a box across the room with a force of 30 N. It accelerates at 5 m/s/s. What is the mass of the box? *

liraira [26]

Given:

Force (F) acting on the box = 30 N

Acceleration produced (a) = 5 m/s2

To determine:

The mass (m) of the box

Calculation:

Formula: F = ma

m = F/a = 30/ 5 = 6 kg

Ans: Mass of the box is 6 kg

5 0

3 years ago

Read 2 more answers