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dusya [7]
3 years ago
13

What is the mass,in grams, of 2.75 L of CO2 gas

Chemistry
1 answer:
frez [133]3 years ago
3 0

Answer:

mass of CO₂ = 5.402 g

Explanation:

At STP, 1 mole of any gas takes up 22.4L of volume

Thus, 1 mole = 22.4 L

1 L = 1 mole ÷ 22.4

Therefore, 2.75 L = 2.75 mole ÷ 22.4 = 0.123 moles

Molecular weight of CO₂ = 44 g/mole

mass of CO₂ = molecular weight * number of moles = 44 g/mole * 0.123 moles

mass of CO₂ = 5.402 g

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Rashid [163]

Answer:

2) It can be performed in natural settings

4 0
1 year ago
How many moles of h+ ions are present in 2.8 l of 0.25 m hydrobromic acid solution?
Romashka [77]

0.7 mol of H⁺ ions are present in 2.8 l of 0.25 m hydrobromic acid solution.

Hydrobromic acid is a strong acid, we can assume that all acid molecules dissociate completely to yield H+ ions and dissociated anion.

<h3>The equation for the dissociation of HBr :</h3>

                             Hbr <em>(s) →  </em>H⁺ <em>(aq)  </em>+  Br⁻<em> (aq)</em>

<em></em>

moles H⁺ = (2800ml) ( \frac{1L}{1000ML})  (\frac{0.25 mol Hbr}{L}) ( \frac{1 mol H}{1 mol Hbr})

               = 0.7 mol

Therefore, 0.7 mol of H⁺ ions is present.

Learn more about H⁺ ions here:

brainly.com/question/12697532

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4 0
1 year ago
Consider the mechanism. Step 1: A+B↽−−⇀CA+B↽−−⇀C equilibrium Step 2: C+A⟶DC+A⟶D slow Overall: 2A+B⟶D2A+B⟶D Determine the rate la
tatuchka [14]

Answer:

rate = k[A][B] where k = k₂K

Explanation:

Your mechanism is a slow step with a prior equilibrium:

\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]

2. Derive the rate law

Assume k₋₁ ≫ k₂.  

Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.  

In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is  

rate = k[A][B] where k = k₂K

5 0
3 years ago
Cell notation will list each half-reaction:
likoan [24]

Answer:

on each side of the salt bridge, which is represented by a double vertical line

Explanation:

While writing a cell notation, the general convention is; anode || cathode. The anode and the cathode are separated by a double line. The anode is written on the lefthand side while the cathode is written on the righthand side.

The cell notation is a shorthand representation of a cell, hence any electrochemical cell can easily be produced based on its cell diagram.

6 0
3 years ago
What is a particle diagram to illustrate the composition of air
navik [9.2K]
A diagram of the composition of air would be the air's chemical formula and it's structure. 
4 0
3 years ago
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