Complete the blank with the correct answer choice. 3 meters = ________ millimeters A. 30 B. 300 C. 3,000 D. 30,000

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

What does the formula H2O mean

forsale [732]

2 hydrogen and 1 oxygen makes water

3 0

3 years ago

Read 2 more answers

What is the frequency of a wave that has 3 waves every five seconds?

Dmitry [639]

Answer:

Frequency, f = 0.6 Hz

Explanation:

We have,

Number of waves passing through a point are 3

Time for which the waves are passing is 5 seconds

It is required to find the frequency of a wave. The frequency of a wave is defined as the no of waves per unit time. So,

$f=\dfrac{n}{t}\\\\f=\dfrac{3}{5}\\\\f=0.6\ Hz$

So, the frequency of a wave is 0.6 Hz.

7 0

3 years ago

PLLLLLLLZ HELP IS NEEDED ASAP I WOULD REALLY APPRECIATE IT

Lesechka [4]

Answer:

1)Reactants

2)Light

3)An item that can increase reaction rates

4)Reactants must collide with each other

Less molecules lower the chance for collisions

The more collisions there are the higher the reaction rate

5 0

3 years ago

600 s after initiation of a first order reaction 48.5% of the initial reactant concentration remains present. What is the rate c

Ludmilka [50]

Answer:

$k=1.20x10^{-3} s^{-1}$

Explanation:

For a first order reaction the rate law is:

$v=\frac{-d[A]}{[A]}=k[A]$

Integranting both sides of the equation we get:

$\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt$

where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.

From that integral we get the integrated rate law:

$ln\frac{[A]}{[A]_{0} } =-kt$

$[A]=[A]_{0}e^{-kt}$

$ln[A]=ln[A]_{0} -kt$

$k=\frac{ln[A]_{0}-ln[A]}{t}$

therefore k is

$k=\frac{ln1-ln0,485}{600}=1,20x10^{-3}$

8 0

3 years ago