Answer:
349.22°C
Explanation:
Let the final temperature of the two pieces of metal be x.
Now, the warmer metal which is C u reduces from 475°C to x. Thus Δt for C u is; Δt1 = 475 - x.
The cooler metal Cr increases in temperature from 265°C to x. Thus, it's change in temperature is Δt for Cr is; Δt2 = x - 265.
Now from conservation of energy, the amount of energy leaving the C u metal is equal to the amount of energy entering the Cr metal.
Thus;
q_lost = q_gain
Where;
q_lost = m1•c1•Δt1
q_gained = m2•c2•Δt2
Now, c1 & c2 are the specific heat capacity of C u and Cr respectively.
From online tables, c1 = 0.385 J/g°C and c2 = 0.46 J/g°C
We are given;
m1 = 12g and m2 = 15g
Thus;
12 × 0.385 × (475 - x) = 15 × 0.46 × (x - 265)
2194.5 - 4.62x = 6.9x - 1828.5
6.9x + 4.62x = 2194.5 + 1828.5
11.52x = 4023
x = 4023/11.52
x = 349.22°C
Answer:
47.5 g of water can be formed
Explanation:
This is the reaction:
CH₄ + 2O₂ → CO₂ + 2H₂O
Methane combustion.
In this process 1 mol of methane react with 2 moles of oxygen to produce 2 moles of water and 1 mol of carbon dioxide.
As ratio is 1:2, I will produce the double of moles of water, with the moles of methane I have.
1.320 mol .2 = 2.64 moles
Now, we can convert the moles to mass (mol . molar mass)
2.64 mol . 18g/mol = 47.5 g
I don't exactly know but probably so.
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:


where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>

- half-life length
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>

</span>




It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>

</span><span>

</span>

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
Answer:
Concentration, because the amounts of reactants and products remain constant after equilibrium is reached.
Explanation:
The rate of reaction refers to the amount of reactants converted or products formed per unit time.
As the reaction progresses, reactions are converted into products. This continues until equilibrium is attained in a closed system.
When equilibrium is attained, the rate of forward reaction is equal to the rate of reverse reaction, hence the concentration of reactants and products in the system remain fairly constant over time.
When deducing the rate of reaction, concentration of the specie of interest is plotted on the y-axis against time on the x-axis.