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MakcuM [25]
3 years ago
8

What is the density of a sample with a mass of 5250g and a volume of 4.5mL

Chemistry
1 answer:
vlabodo [156]3 years ago
5 0

Answer:

1166.67

Explanation:

density=mass/volume

density= 5250/4.5

density=1166.67g/mL

You might be interested in
Help me please an thank you
miskamm [114]

Molecular formula

hope that helps^^

6 0
3 years ago
95. Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulfur double bond in C
prohojiy [21]

Answer:

+523 kJ.

Explanation:

The following data will be used to calculate the average C-S bond energy in CS2(l).

S(s) ---> S(g)

ΔH = 223 kJ/mol

C(s) ---> C(g)

ΔH = 715 kJ/mol

Enthalpy of formation of CS2(l)

ΔH = 88 kJ/mol

CS2(l) ---> CS2(g)

ΔH = 27 kJ/mol

CS2(g) --> C(g) + 2S(g)

So we must construct it stepwise.

1: C(s) ---> C(g) ΔH = 715 kJ

2: 2S(s) ---> 2S(g) ΔH = 446 kJ

adding 1 + 2 = 3

ΔH = 715 + 446

= 1161 kJ

3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ

4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ

adding (reversed 3) from 4 = 5

ΔH = -1161 + 88

= -1073 kJ

5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ

6: CS2(l) ---> CS2(g) ΔH = 27 kJ

adding 5 + 6 = 7

ΔH = -1073 + 27

= -1046 kJ

7. C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ

Reverse and divide by 2 for C-S bond enthalpy

= -(-1046)/2

= +523 kJ.

8 0
3 years ago
Vitamin C has the formula CxHyOz. You burn 0.486 g of the compound in a combustion analysis chamber and isolate 0.728 g of CO2 a
-Dominant- [34]

Answer:

C_3H_4O_3

Explanation:

Hello there!

In this case, according to the given information derived from the combustion analysis, it turns out possible for us to realize that all the carbon comes from the CO2 and all the hydrogen from the H2O, it means we can calculate their moles in the vitamin C as shown below:

n_C=0.728gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.0165molC\\\\n_H=  0.198gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.0220molH

Next, we calculate the grams and moles of oxygen from the grams of C and H in the sample:

m_O=0.486g-0.0165molC*\frac{12.01gC}{1molC} -0.0220molH*\frac{1.01gH}{1molH}\\\\m_O=0.266gO\\\\n_O=0.266gO*\frac{1molO}{16.00gO} =0.0166molO

Then, we divide the moles of C, H, O by 0.0165 as the fewest moles in order to calculate the correct mole ratios:

C=\frac{0.0165}{0.0165}= 1\\\\H=\frac{0.0220}{0.0165}= 1.33\\\\O=\frac{0.0166}{0.0165}= 1

Finally, we turn them into whole number by multiplying by 3 so that the empirical formula is:

C_3H_4O_3

Regards!

3 0
3 years ago
Read 2 more answers
Kathy has been exercising vigorously for the past half hour. her body feels sore and she feels pain in her limbs. what could be
valkas [14]
I believe it is A accumulation of lactic acid in the muscles of her limbs.
5 0
3 years ago
Read 2 more answers
What does this do to the electrons outside the nucleus in the gaseous atoms
AleksandrR [38]

Answer:

Explanation:

As you know, ionization energy is the energy needed to remove one mole of electrons from one mole of atoms in the gaseous state

X

+

energy

→

X

+

+

e

−

Right from the start, you can tell that the harder it is to remove an electron from an atom, the higher the ionization energy will be.

Now, the periodic trends for ionization energy can be describe as follows

ionization energy increases as you move from left to right across a period

ionization energy decreases as you go down a group

As you mentioned, if you compare the first ionization energies for oxygen and chlorine using these two trends, you will get conflicting results.

If you follow the way ionization energy increases across period, chlorine would have a higher ionization energy, since it's closer to the noble gases.

On the other hand, if you go by how ionziation energy decreases from top to bottom in a group, oxygen would have higher ionization energy, since it's located in period 2, as compared with period 3 for chlorine.

As it turns out, the trend for groups overpowers the trend for periods. As aresult, oxygen will have a higher ionization energy than chlorine.

This happens because the smaller oxygen atom has its outermost electrons held tighter by the nucleus. By comparison, chlorine's outermost atoms are located further away from the nucleus.

Not only that, but they are screened from the charge of the nucleus better, since they're located on the third energy level.

Oxygen's outermost electrons are screened by

2

electrons, while chlorine's are screened by

8

electrons.

All these factors will make chlorine's outermost electrons a little easier to remove, which implies a smaller ionization energy than that of oxygen.v

6 0
3 years ago
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