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3 years ago

How should the enthalpy of an intermediate step be manipulated when used to produce an overall chemical equation?

1 answer:

8 0

The enthalpy of an intermediate step should be manipulated when used to produce an overall equation by using the Hess's law. You could multiply the enthalpy by -1 if this equation is reversed in theory.

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What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Vadim26 [7]

Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
c= 0.4494 J/g°C
ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$

The mass of the sample of metal is approximately <u>333 grams.</u>

8 0

2 years ago

an element in group two has two valence electrons. When forming a chemical bond it will most likely (gain, lose, share, or not r

Charra [1.4K]

Lose electrons - electrons want to fill their outer valence shell, so sometimes instead of gaining it is easier to lose some and have a filled outer shell

3 0

3 years ago

The burr oak trees in the forest are which of these?

hammer [34]

Answer:

I think It is an organism

4 0

2 years ago

Read 2 more answers

1s22s22p63s23p6 How many unpaired electrons are in the atom represented by the electron configuration above?

pickupchik [31]

Answer:

Explanation:

There are no unpaired electrons in the given element. It must be noted that for the atom above, we have even numbered electrons. The total electron we are having here is 18.

Now, we must also know that while the s orbital is not degenerate, the P orbital is degenerate. What this mean is that the p orbital is broken down into three different sub orbitals which is the Px , Py and Pz. Hence we can see that there are 6 electrons to enter into the P orbital too.

We can see that all the S orbitals have been completely filled with two electrons alike each. This is also the case for the P orbital as the 3 suborbitals take in 2 each to give a total of six

4 0

3 years ago

Which type of orbits are found in the principal energy level n = 2

Vilka [71]

D. s, p is your answer.

4 0

3 years ago