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Is a Big Mac an element, compound, or mixture?

Lunna [17]

It is a mixture. it can easily be separated.

4 0

4 years ago

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

Although I am solid, I am so light that I can float in any liquid listed on the chart. What am I?-----

evablogger [386]

Answer:

I know that Aerogel is the lightest metal in existence, but I don't think it would help much with your answer. I mean you can give it a try?

3 0

3 years ago

Chlorine (Cl) is 76% chlorine-35 and 24% chlorine-37. Determine the average atomic mass of chlorine. Show/Explain work.

Flura [38]

Answer:

Average atomic mass of chlorine is 35.48 amu.

Explanation:

Given data:

Percent abundance of Cl-35 = 76%

Percent abundance of Cl-37 = 24%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100

Average atomic mass = (76×35)+(24×37) /100

Average atomic mass = 2660 + 888 / 100

Average atomic mass = 3548/ 100

Average atomic mass = 35.48 amu

Average atomic mass of chlorine is 35.48 amu.

4 0

3 years ago

What is the explanation as to why bonds form

VikaD [51]

<span>They want a full outer shell of electrons, so the lose, gain, or share electrons with other elements, forming compounds, until they have 8 valence electrons and become stable.

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4 0

3 years ago

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