Answer:
B. CaCl₂ + H₂CO₃ → CaCO₃ + 2HC
Explanation:
A balanced reaction has the same number of atoms in the both sides of the reaction. In the options:
A. CaCl₂ + H₂CO₃ → 2CaCO₃ + HCI
In this reaction there is 1 Ca in reactants and 2 in products -<em>The reaction is unbalanced-</em>
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<h3>B. CaCl₂ + H₂CO₃ → CaCO₃ + 2HCl
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There is 1 Ca is both sides, 2Cl, 2H, 1C and 3 Oxygens -<em>The reaction is balanced</em>
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C. CaCl₂ + 2H₂CO₃ → CaCO₃ + HCI
There is 1 Ca in both sides but 2Cl in reactants and 1 in Cl -<em>The reaction is unbalanced-</em>
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D. 2CaCl₂ + H₂CO₃ →CaCO₃ + HCI
There are 2 Ca in reactants and 1 in Ca -<em>The reaction is unbalanced-</em>
<h3>
Answer:</h3>
Anion present- Iodide ion (I⁻)
Net ionic equation- Ag⁺(aq) + I⁻(aq) → AgI(s)
<h3>
Explanation:</h3>
In order to answer the question, we need to have an understanding of insoluble salts or precipitates formed by silver metal.
Additionally we need to know the color of the precipitates.
Some of insoluble salts of silver and their color include;
- Silver chloride (AgCl) - white color
- Silver bromide (AgBr)- Pale cream color
- Silver Iodide (AgI) - Yellow color
- Silver hydroxide (Ag(OH)- Brown color
With that information we can identify the precipitate of silver formed and identify the anion present in the sample.
- The color of the precipitate formed upon addition of AgNO₃ is yellow, this means the precipitate formed was AgI.
- Therefore, the anion that was present in the sample was iodide ion (I⁻).
- Thus, the corresponding net ionic equation will be;
Ag⁺(aq) + I⁻(aq) → AgI(s)
Answer:
Q = 306 kJ
Explanation:
Given that,
Mass, m = 60 kg
Specific heat, c = 1020 J/kg°C
The temperature changes from 20°C to 25°C.
Let Q be the change in thermal energy. The formula for the heat released is given by :
Put all the values,
So, 306 kJ is the change in thermal energy.
Answer : The final temperature of the metal block is,
Explanation :
As we know that,
.................(1)
where,
q = heat absorbed or released
= mass of aluminum = 55 g
= mass of water = 0.48 g
= final temperature = ?
= temperature of aluminum =
= temperature of water =
= specific heat of aluminum =
= specific heat of water=
Now put all the given values in equation (1), we get
Thus, the final temperature of the metal block is,
I believe it’s the last option