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VLD [36.1K]
3 years ago
9

What is the volume, in milliliters, occupied by 30.07 g of an object of density equal to

Chemistry
1 answer:
den301095 [7]3 years ago
3 0

Answer:

\boxed {\tt 20.317567567568 \ mL}

Explanation:

The density formula is:

d=\frac{m}{v}

Let's rearrange the formula for v. the volume. Multiply both sides by v, then divide by d.

d*v=\frac{m}{v}*v

d*v=m

\frac{d*v}{d}=\frac{m}{d}

v=\frac{m}{d}

The volume can be found by dividing the mass by the density. The mass of the object is 30.07 grams and the density is 1.48 grams per milliliter.

m= 30.07 \ g\\d= 1.48 \ g/mL

v=\frac{30.07 \ g}{1.48 \ g/mL}

Divide. Note, when dividing, the grams, or g will cancel out.

v= \frac{30.07}{1.48 \ mL}

v=20.317567567568 \ mL

The volume of the object is 20.317567567568 milliliters.

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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
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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}

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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.

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