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Leokris [45]
3 years ago
15

A copper rod that has a mass of 200.0 g has an initial temperature of 20.0°C and is heated to 40.0°C. If 1,540 J of heat are nee

ded to heat the rod, what is the specific heat of copper?
Chemistry
2 answers:
Romashka [77]3 years ago
8 0

Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)</span>

1540 = 200.0 (C)(40 - 20)

<span> <span>C = 0.385 J / g C</span></span>

<span><span>
</span></span>

<span><span>Hope this answers the question. Have a nice day.</span></span>

Molodets [167]3 years ago
5 0

Answer:

D on edge

Explanation:

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Answer:

HCl, also known as hydrochloric acid, has a covalent bond. The hydrogen (H) atom shares an electron with the chlorine (Cl) to form the bond.

Explanation:

Consequently, the bonding electrons in hydrogen chloride are shared unequally in a polar covalent bond. The molecule is represented by the conventional Lewis structure, even though the shared electron pair is associated to a larger extent with chlorine than with hydrogen.

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3 years ago
When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

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the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

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If you have 58.93g of Co it means that you only have 1 mol (use a periodic table to find the answer, if you had more find it by proportion, it's easier). 
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(once again if you had more mol, you could find the answer by proportions).
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