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jek_recluse [69]
3 years ago
14

A solid with a low melting point is most likely

Chemistry
2 answers:
Reil [10]3 years ago
7 0
A solid with a low melting point is most likely is held together by covalent bonds. Examples are hydrocarbons, ice, sugar and sulfur. The have low melting points because of the covalent bonds .  It do not form crystals therefore can easily be broken. The attractions are weak. 
mezya [45]3 years ago
7 0

Answer:

Necesito un cuento sobre romance

Explanation:

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A lead mass is heated and placed in a foam cup calorimeter containing 50.0 ml of water at 18.0°c. the water reaches a temperatur
garri49 [273]
Heat gained or loss 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) 

When two objects are in contact, it should be that the heat lost is equal to what is gained by the other. So, the heat released by the lead is equal to the heat that is absorbed by the water. 

</span>Heat = mC(T2-T1) = 50.0 mL (1.00 g/mL) (4.18 J/g °C) (20 °C - 18 °C) = 418 J<span>

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8 0
3 years ago
Read 2 more answers
What are 5 renewable energy Resources and why are they better for the society and the environment
Ket [755]

Answer:

hydroelectric , hydrogen fuel cells , solar power , geothermal , wind power

Explanation:

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5 0
3 years ago
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---&gt; +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

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PE, GO, XY - I am  probably wrong xoxoxoxoxxo
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Gasoline burning in an engine: Exothermic or Endothermic?
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