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amm1812
3 years ago
10

The specific heat capacity of copper is 0.385 J/(gK). If 353 J of heat are added to 3.6 moles of copper at 283 K, what is the fi

nal temperature of the sample of copper?
Chemistry
1 answer:
hram777 [196]3 years ago
6 0

Answer:

287K

Explanation:

To find the increasing in temperature for the sample of copper we must use the equation:

Q = m*C*ΔT

<em>Q is the heat added (353J)</em>

<em>m is the mass of copper (3.6mol * (63.546g / mol) = 228.8g </em>

<em>C is specific heat (0.385J/(gK))</em>

<em>And ΔT is the increasing in temperature.</em>

<em />

Solving for ΔT:

Q / m*C = ΔT

353J /228.8g*0.385J/(gK) = 4.0K

That means the final temperature of copper is:

283K + 4K =

<h3>287K</h3>
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Answer:

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Covalent Bonding: Bonding between non-metals consists of two electrons shared between two atoms.

Explanation:

4 0
3 years ago
Which of the following equations does not demonstrate the law of conservation of mass?
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The third option does not obey the law of conservation of mass.

Option 3.

Explanation:

The law of conservation of mass states that the sum of the masses of reactants should be equal to the sum of the masses of the products.

For example, if we consider the first option to verify if it obeys law of conservation of mass or not, 2 Na + Cl₂ → 2 NaCl

So one way to verify it is to find the mass of Na, then multiply it with 2, and then add this with 2 times of mass of chlorine. So this sum should be equal to the 2 times mass of NaCl. But it is somewhat lengthy.

Another way to easily determine this is to check if the elements are present equally in both sides. Such as, in reactant side and product side 2 atoms of Na is present . Similarly, the Cl atoms are also present in equal number in both reactant and product side. Thus this obeyed the law of conservation of mass.

Like this, if we see the second option, there also 1 atom of Na is present in reactant and product side and 2 molecules of H is present in reactant and product side, 1 oxygen is present in reactant and product side and 1 Cl is present in reactant and product side. So it also obeys the law of conservation of mass.

But in the third option, P₄ + 5 O₂→ 2 P₄O₁₀, here, there is 4 atoms of P in reactant side but in product side there is (4*2) = 8 atoms of P. Similarly, the number of atoms of oxygen in reactants and product side is also not same. So the third option does not obey the law of conservation of mass.

The fourth option also obeys the law of conservation of mass as the number of atoms of each element is same in both the product and reactant side.

Thus, the third option does not obey the law of conservation of mass.

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Ivahew [28]

Answer:

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In order to convert from grams of any given substance to moles, we need to use its molar mass:

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Now we <u>calculate the number of moles of KAI(SO₂)₂ contained in 5.98 g</u>:

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Thus, the correct answer is option (D).

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3 years ago
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