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Andre45 [30]
3 years ago
5

One tablespoon of peanut butter has a mass of 15 g . It is combusted in a calorimeter whose heat capacity is 120 kJ/∘C . The tem

perature of the calorimeter rises from 22.4 ∘C to 25.5 ∘C. Find the food caloric content of peanut butter.
Chemistry
1 answer:
devlian [24]3 years ago
8 0

Answer:

5927.3488 calorie/g

Explanation:

For peanut butter:

Mass = 15 g

Initial temperature = 22.4 °C

Final temperature = 25.5 °C

Specific heat of calorimeter, C = 120 kJ/°C

Considering

Q=C\times (T_f-T_i)

Q=120\ kJ/^0C\times (25.5-22.4)\ ^0C

Q=372\ kJ

Also,

1 kJ = 239.006 calories

So, Heat = 372*239.006 calories = 88910.232 calories

15 g of the peanut butter gives 88910.232 calories

<u>So, the food caloric content of peanut butter = 88910.232 calories/ 15 g = 5927.3488 calorie/g</u>

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Water is polar because of the bent shape of the molecule.

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Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.

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Carbon is a chemical element with symbol C and atomic number 6, which means it has 6 protons and six electrons. Four valence electrons are in 2s and 2p orbitals.  

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5 0
3 years ago
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Explanation:

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4 0
3 years ago
The table shows the number of reactants and products present during two separate chemical reactions.
finlep [7]

Answer:

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Explanation:

A synthesis or combination reaction is that reaction that involves combination of two or more reactants to form a single product. The two or more reactants are often elements that chemically combine to form a single compound.

In this question, two chemical reactions are involved in which chemical reaction A has 1 reactants and 3 products while chemical reaction B has 2 reactants and 1 product. Based on the explanation above, the reaction B correctly identifies the synthesis reaction because the single product is a compound.

4 0
3 years ago
Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r
Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

7 0
3 years ago
Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
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