0.208 is the specific heat capacity of the metal.
Explanation:
Given:
mass (m) = 63.5 grams 0R 0.0635 kg
Heat absorbed (q) = 355 Joules
Δ T (change in temperature) = 4.56 degrees or 273.15+4.56 = 268.59 K
cp (specific heat capacity) = ?
the formula used for heat absorbed and to calculate specific heat capacity of a substance will be calculated by using the equation:
q = mc Δ T
c = 
c = 
= 0.208 J/gm K
specific heat capacity of 0.208 J/gm K
The specific heat capacity is defined as the heat required to raise the temperature of a substance which is 1 gram. The temperature is in Kelvin and energy required is in joules.
Answer : The enthalpy change during the reaction is -6.48 kJ/mole
Explanation :
First we have to calculate the heat gained by the reaction.
where,
q = heat gained = ?
m = mass of water = 100 g
c = specific heat = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:
Now we have to calculate the enthalpy change during the reaction.
where,
= enthalpy change = ?
q = heat gained = 23.4 kJ
n = number of moles barium chloride = 
Therefore, the enthalpy change during the reaction is -6.48 kJ/mole
Answer
D 160g
Explanation:
<u>Write the equation:</u>
Combustion reactions use oxygen and release water and heat, so
CH₃OH(g) + O₂(g) → CO₂(g) + H₂O(g)
Balance that:
2CH₃OH(g) + 3O₂(g) → 2CO₂(g) + 4H₂O(g)
<u>Find moles of carbon dioxide:</u>
We need to know the number of moles of CO₂. This rxn is at STP, so at STP one mole of gas = 22.4 liters.
112 L * 1 mol/22.4 L = <em>5 mol CO₂</em>
<u>Find moles of methanol:</u>
Based on the chemical equation, for every 2 mol methanol, there are 2 mol carbon dioxide. So for every 5 mol carbon dioxide, there are 5 mol methanol!
5 mol CO₂ = 5 mol CH₃OH
Molar mass of methanol: 12.01 + 3*1.008 + 16.00 + 1.008 = <em>32.04 g/mol</em>
Moles of methanol: 5 mol * 32.04 g/mol = 160.2 g methanol
≈ 160 mol methanol
Answer:
The pH of the buffer is 7.0 and this pH is not useful to pH 7.0
Explanation:
The pH of a buffer is obtained by using H-H equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pH is the pH of the buffer</em>
<em>The pKa of acetic acid is 4.74.</em>
<em>[A⁻] could be taken as moles of sodium acetate (14.59g * (1mol / 82g) = 0.1779 moles</em>
<em>[HA] are the moles of acetic acid (0.060g * (1mol / 60g) = 0.001moles</em>
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Replacing:
pH = 4.74 + log [0.1779mol] / [0.001mol]
<em>pH = 6.99 ≈ 7.0</em>
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The pH of the buffer is 7.0
But the buffer is not useful to pH = 7.0 because a buffer works between pKa±1 (For acetic acid: 3.74 - 5.74). As pH 7.0 is out of this interval,
this pH is not useful to pH 7.0
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