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ELEN [110]
3 years ago
5

What kind of solid tends to have the lowest melting points?

Chemistry
1 answer:
pishuonlain [190]3 years ago
6 0

Answer:

Molecular solids

Covalent compounds

Explanation:

vote me as the brainliest

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There are two common oxides of copper; one is 80% copper and the other is 89% copper by weight. Calculate the formula and name e
irakobra [83]

80% copper (Cu)

Cu: 80 :  63.546 = 1.259

O: 20 : 16 = 1.25

Cu:O = 1 : 1

the formula: CuO

89% copper (Cu)

Cu: 89 :  63.546 = 1.4

O: 11 : 16 = 0.6875

Cu:O = 2:1

the formula: Cu₂O

8 0
2 years ago
compound 1 contains 15.0g of hydrogen and 120.0g oxygen. What is the percent compound of each element?
allsm [11]

Note down the formula below

\boxed{\sf Mass\%\;of\; element=\dfrac{Mass\:of\:the\: element}{Mass\;of\:the\: compound}\times 100}

Mass of the compound

\\ \sf\longmapsto 15+120=135g

Mass % of Hydrogen:-

\\ \sf\longmapsto \dfrac{15}{135}\times 100

\\ \sf\longmapsto \dfrac{1}{9}\times 100

\\ \sf\longmapsto 11.1\%

Mass % of Oxygen:-

\\ \sf\longmapsto \dfrac{120}{135}\times 100

\\ \sf\longmapsto \dfrac{8}{9}\times 100

\\ \sf\longmapsto 88.9\%

8 0
3 years ago
The enthalpy change for converting 1.00 mol of ice at -50.0°c to water at 70.0°c is __________ kj. the specific heats of ice, wa
kogti [31]
First, calculate for the amount of heat used up for increasing the temperature of ice.

      H = mcpdT
       H = (18 g)*(2.09 J/g-K)(50 K) = 1881 J

Then, solve for the heat needed to convert the phase of water.
    H = (1 mol)(6.01 kJ/mol) = 6.01 kJ = 6010 J

Then, solve for the heat needed to increase again the temperature of water.
    H = (18 g)(4.18 J/gK)(70 k)
    H = 5266.8 J

The total value is equal to 13157.8 J

Answer: 13157.8 J
8 0
3 years ago
What is the basic unit of chemistry?
Luba_88 [7]
An atom hopefully this helps
7 0
3 years ago
Read 2 more answers
When 1.50 g of ba(s) is added to 100.00 g of water in a container open to the atmosphere, the reaction shown below occurs and th
damaskus [11]

<u>Given:</u>

Mass of Ba = 1.50 g

Mass of H2O = 100.0 g

Initial temp T1 = 22 C

Final Temp T2 = 33.1 C

specific heat c = 4.18 J/g c

<u>To determine:</u>

The reaction enthalpy

<u>Explanation:</u>

The heat released during the reaction is:

q = - mc(T2-T1) = - (100+1.5) g *4.18 J/g C * (33.1-22) C = -4709.4 J

# moles of Ba = Mass of Ba/Atomic mass of Ba = 1.5 g/137 g.mol-1 = 0.0109 moles

ΔH = q/mole = - 4709.4 J/0.0109 moles = - 432 kJ/mol

Ans : The enthalpy change for the reaction is -432 kJ/mol


4 0
3 years ago
Read 2 more answers
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