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2 years ago

Diluting sulfuric acid with water is highly exothermic:

Chemistry

1 answer:

elena-14-01-66 [18.8K]2 years ago

8 0

Answer:

The correct answer is 51.2 degree C.

Explanation:

The standard enthalpy for H₂SO₄ (l) is -814 kJ/mole and the standard enthalpy for H₂SO₄ (aq) is -909.3 kJ/mole.

Now the dHreaction = dHf (product) - dHf (reactant)

= -909.3 - (-814)

dHreaction or q = -95.3 kJ of energy will be used for dissociating one mole of H₂SO₄.

The heat change in calorimetry can be determined by using the formula,

q = mass * specific heat capacity * change in temperature -----------(i)

Based on the given information, the density of H₂SO₄ is 1.060 g/ml

The volume of H₂SO₄ is 1 Liter

Therefore, the mass of H₂SO₄ will be, density/Volume = 1.060 g/ml / 1 × 10⁻³ ml = 1060 grams

The initial temperature given is 25.2 degrees C, or 273+25.2 = 298.2 K, let us consider the final temperature to be T₂.

ΔT = T₂ -T₁ = T₂ - 298.2 K

Now putting the values in equation (i) we get,

95.3 kJ = 1060 grams × 3.458 j/gK (T₂ - 298.2 K) (the specific heat capacity of the final solution is 3.458 J/gK)

(T₂ - 298.2 K) = 95300 J / 1060 × 3.458 = 26 K

T₂ = 298.2 K + 26 K

T₂ = 324.2 K or 324.2 - 273 = 51.2 degree C.

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Answer:

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2 years ago

The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha

zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$25.9 = x*10.0$

$25.9 = 10.0\:x$

$10.0\:x = 25.9$

$x = \dfrac{25.9}{10.0}$

$\boxed{x = 2.59}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{53.3}{2^{2.59}}$

$m \approx \dfrac{53.3}{6.021}$

$\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

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3 years ago

8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a

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212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

$C_{1} V_{1} =C_{2} V_{2}$