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polet [3.4K]
3 years ago
15

Crude oil may contain hundreds of different types of hydrocarbons. Some examples include: Butane(C4H10) Dodecane (C12H26) Octane

(C8H18) Benzene (C6H6) Many common fuels, such as gasoline and kerosene, are combinations of these substances or others. When these fuels burn, they combine with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
Chemistry
1 answer:
Nesterboy [21]3 years ago
7 0

Answer:

Butane and benzene are both compounds. They contain multiple elements bonded together in a specific ratio. Kerosene and gasoline are mixtures because they are combinations of several compounds. Oxygen is an element because it is made up of only one type of atom.

Explanation:

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A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0
miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

Q_{metal} - Q_{water} = 0

Q_{metal} = Q_{water}

Where:

Q_{water} - Heat received by water, measured in joules.

Q_{metal} - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})

The specific heat of the metal is cleared within equation above:

c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}

If we know that m_{water} = 0.237\,kg, m_{metal} = 0.244\,kg, c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}, T_{w,o} = 10\,^{\circ}C, T_{m,o} = 100.10\,^{\circ}C and T = 15.30\,^{\circ}C, the specific heat of molybdenum is:

c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}

c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

5 0
3 years ago
Graphite and diamond are both solid forms of the element carbon. Which statement explains
vlada-n [284]

Answer: C diamond has a different crystal structure from graphite

Explanation:

3 0
2 years ago
When ammonium phosphate and nickel(ii) nitrate solutions are mixed, what precipitates out?
Sphinxa [80]

2(NH4)3PO4 (aq) + 3Ni(NO3)2(aq) ------> Ni3(PO4)2(s) + 6NH4NO3 (aq)

Ni3(PO4)2 is a precipitate.

5 0
3 years ago
Read 2 more answers
What is a chemical off of the periodic table that is sticky
Alex73 [517]
I think it’s the one that has a Br
3 0
3 years ago
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How many grams of tris (mw 121.1) would you need to prepare 100 ml of a 100 mm tris solution? _________ grams?
Tems11 [23]

Molarity = (Mass/ molar mass) x (1/ volume of solution in Litres)

Mass = Molarity x molar mass x  volume of solution in Litres

Molarity of Tris = 100 mM = 0.1 M

volume of Tris sol. = 100 mL = 0.1 L

molar mass of Tris = 121.1 g/mol

Hence,

mass of Tris = Molarity of Tris x molar mass ofTris x volume of Tris solution

= 0.1 M x 121.1 g/mol x 0.1 L

= 1.211 g

mass of Tris = 1.211 g


7 0
3 years ago
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