The correct answer is 3) 2CO2(g) ⇄ 2CO(g) + O2(g)
this is the correct one because it is a decomposition reaction and all the number of atoms is equal on both sides.
there are 2 C atoms on both sides.
and 4 O atoms on both sides.
and 1) the atoms numbers are equal on both sides but not correct as it not a
correct number as it has 1/2 O2.
and 2) CO2(g) ⇆ CO(g) + O2
the number of O atoms is not equal on both sides of the equation.
we have 2 O atoms on the left side and 3 O atoms on the right side.
so, this not a balanced equation.
4) also not correct 2CO(g) + O2 ⇆ 2CO2
as it is not a decomposition reaction and the 2CO & O2 are as reactants not products.
so the correct answer is 3) 2CO2(g) ⇆ 2CO(g) + O2(g)
<h3>
Answer:</h3>
The centripetal acceleration is 26.38 m/s²
<h3>
Explanation:</h3>
We are given;
- Mass of rubber stopper = 13 g
- Length of the string(radius) = 0.93 m
- Time for one revolution = 1.18 seconds
We are required to calculate the centripetal acceleration.
To get the centripetal acceleration is given by the formula;
Centripetal acc = V²/r
Where, V is the velocity and r is the radius.
Since time for 1 revolution is 1.18 seconds,
Then, V = 2πr/t, taking π to be 3.142 ( 1 revolution = 2πr)
Therefore;
Velocity = (2 × 3.142 × 0.93 m) ÷ 1.18 sec
= 4.953 m/s
Thus;
Centripetal acceleration = (4.953 m/s)² ÷ 0.93 m
= 26.38 m/s²
Hence, the centripetal acceleration is 26.38 m/s²
Answer:
0.0277 M.
Explanation:
The integral rate law of a first order reaction:
<em>Kt = ln ([A₀]/[A]),</em>
where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,
t is the time of the reaction <em>(t = 235.0 min = 14100 s)</em>,
[A₀] is the initial concentration of cyclopropane <em>([A₀] = 0.0445 M)</em>
<em>∵ Kt = ln ([A₀]/[A]),</em>
∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]
Taking the exponential of both sides:
1.6 = (0.0445 M)/[A]
<em>∴ [A] = (0.0445 M)/1.6 = 0.0277 M.</em>
<em />
40.1g of nitrogen gas is produced.
The equation given is
2 NH₃ + 3 CuO →3 Cu + N₂ + 3 H₂O
This equation is already balanced.
When 3 moles of CuO are consumed, 1 mole of nitrogen gas is produced.
We get 1 mole of nitrogen from 3 moles of copper oxide.
We need to find the number of moles of nitrogen gas produced when 4.3 moles of copper oxide are consumed.
4.3/3 x 1 = 1.433 mols
- 1.433 mols of nitrogen gas are produced
- The molar mass of nitrogen gas is 14+14 = 28g
- The amount of nitrogen gas produced in grams is 28x1.433 = 40.1g
40.1g of nitrogen gas can be made when 4.3 moles of CuO are consumed.
Learn more about molarity here:
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600,000 mm if im not mistaken.
