<span>We can use
the heat equation,
Q = mcΔT </span>
<span>Where Q is
the amount of energy transferred (J), m is the mass of the substance
(kg), c is
the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is
the temperature difference (°C).</span>
Let's assume that the finale temperature is T.
Q = 1200 J
<span>
m = 36 g
c = 4.186 J/g °C</span>
ΔT = (T -
22)
By applying
the formula,
1200 J = 36 g
x 4.186 J/g °C x (T - 22)
(T - 22) = 1200 J / (36 g x 4.186 J/g °C)
(T - 22) = 7.96 °C
T = (7.96 + 22) °C = 29.96 °C
T = 30 °C
Hence,
the final temperature is 30 °C.
(a) In this section, give your answers to three decimal places.
(i)
Calculate the mass of carbon present in 0.352 g of CO
2
.
Use this value to calculate the amount, in moles, of carbon atoms present in 0.240 g
of
A
.
(ii)
Calculate the mass of hydrogen present in 0.144 g of H
2
O.
Use this value to calculate the amount, in moles, of hydrogen atoms present in 0.240 g
of
A
.
(iii)
Use your answers to calculate the mass of oxygen present in 0.240 g of
A
Use this value to calculate the amount, in moles, of oxygen atoms present in 0.240 g
of
A
(b)
Use your answers to
(a)
to calculate the empirical formula of
A
thank you
hope it helpsss
Answer:
See explaination
Explanation:
please kindly see attachment for the step by step solution of the given problem.
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
c₁=2.00 mol/L
v₁=0.25 L
v₂=2.00 L
c₂-?
n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
n(NaOH)=2n(H₂SO₄)
c₂v₂=2c₁v₁
c₂=2c₁v₁/v₂
c₂=2*2.00*0.25/2.00=0.5 mol/L
0.5 M NaOH
Answer:
How can you distinguish a physical change from a chemical change?
Explanation:
