Protein=Amino acid
Carbohydrates=glucose
FATS=LIPIDS
Answer:

Explanation:
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In this case, according to the reaction:

We can see there is a 4:2 mole ratio between chromium and chromium (III) oxide, this, for the given 56.2 g of chromium, the theoretical yield of the oxide product is computed down below:

Now, considering the 76.0-% yield for this reaction, the actual yield turns out:

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At equivalence there is no more HA and no more NaOH, for this particular reaction. So that means we have a beaker of NaA and H2O. The H2O contributes 1 x 10-7 M hydrogen ion and hydroxide ion. But NaA is completely soluble because group 1 ion compounds are always soluble. So NaA breaks apart in water and it just so happens to be in water. So now NaA is broken up. The Na+ doesn't change the pH but the A- does change the pH. Remember that the A anion is from a weak acid. That means it will easily attract a hydrogen ion if one is available. What do you know? The A anion is in a beaker of H+ ions! So the A- will attract H+ and become HA. When this happens, it leaves OH-, creating a basic solution, as shown below.
Answer:
5446.8 J
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 50 g
Initial temperature (T₁) = 70 °C
Final temperature (T₂) = 192.4 °C
Specific heat capacity (C) = 0.89 J/gºC
Heat (Q) required =?
Next, we shall determine the change in the temperature. This can be obtained as follow:
Initial temperature (T₁) = 70 °C
Final temperature (T₂) = 192.4 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 192.4 – 70
ΔT = 122.4 °C
Finally, we shall determine the heat required to heat up the block of aluminum as follow:
Mass (M) = 50 g
Specific heat capacity (C) = 0.89 J/gºC
Change in temperature (ΔT) = 122.4 °C
Heat (Q) required =?
Q = MCΔT
Q = 50 × 0.89 × 122.4
Q = 5446.8 J
Thus, the heat required to heat up the block of aluminum is 5446.8 J