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2 years ago

Gold has a specific heat of 0.126 J/g.C. Copper has a specific heat of 0.386 J/g C. Which of the two metals would require more

Chemistry

1 answer:

ASHA 777 [7]2 years ago

8 0

Answer:

The copper, because its specific heat is higher, meaning it takes more heat (Joules) per gram to raise the temperature 1 degree Celsius.

Explanation:

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Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina

yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

8 0

3 years ago

The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:

aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium: 0.50M - x x x

The equilibrium constant (Ka₁) is:

$K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}$

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

$[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M$

Similarly, from the second dissociation of sulfurous acid we have:

HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium: 7.94x10⁻²M - x x x

The equilibrium constant (Ka₂) is:

$K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}$

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

$[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M$

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0

3 years ago

How many milliliters of .085 m naoh are required to titrate 25 ml of .072 m hbr to the equivalence point?

SOVA2 [1]

The ML of 0.85 m NaOH required to titrate 25 ml of 0.72m hbr to the equivalence point is calculated as follows

calculate the moles of HBr used

moles = molarity x volume

25 x0.072/1000= 0.0018 moles

write the equation for reaction

NaOH + HBr = NaBr + H2O
from reacting equation the mole ratio between NaOH to HBr is 1:1 therefore the moles of NaOH = 0.0018 moles

volume = moles/molarity
0.0018/0.085 = 0.021 L in Ml = 0.021 x1000=21.18 Ml ofNaOH

8 0

3 years ago

A solution...

steposvetlana [31]

Answer:

Explanation:

8 0

3 years ago

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You measure salt water in a tank to have a density of 1.02 g/mL. A balloon weighs 2.0 g and you weights have a mass of 30.0 g ea

Elden [556K]

Weight of the balloon = 2.0 g