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3 years ago

Please help easy chemistry

Chemistry

1 answer:

Rzqust [24]3 years ago

5 0

Answer:

The answer to your question is the third option, atoms of each lose one electron to achieve stability.

Explanation:

a) The first option is incorrect, those elements are extremely reactive and can explode.

b) The option is also incorrect, metalloids are located in groups 3A, 4A OR 5A in the periodic table.

c) This option is correct

d) These elements have 1 valence electron, this option is wrong.

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Plz help me out with this <br>10 points

grin007 [14]

Answer:

you can tell diorite is an intrusive igneous rock because it has a coarse-grained texture

6 0

3 years ago

Read 2 more answers

CH3(CH2)50CH3<br><br> How many atoms

ahrayia [7]

Explanation:

An atom is the smallest indivisible particle of any matter. Atoms are distinct and distinguishable from one another and are called elements:

CH₃(CH₂)₅0CH₃

There are 3 different kinds of atoms in the compound above called

elements

Carbon

Hydrogen

Oxygen

In the compound the subscripts gives the number of atoms:

C - 7

H - 16

0 - 1

Number of atoms in the compound is 24.

Learn more:

Number of atoms brainly.com/question/10419836

#learnwithBrainly

3 0

3 years ago

TIMED

zhannawk [14.2K]

Density = mass/volume
Density = 81g/0.9cm³
Density = 90g/cm³
: )

6 0

3 years ago

Read 2 more answers

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

Consider the reaction: 2 SO2(g)+O2(g)→2 SO3(g) If 285.5 mL of SO2 reacts with 158.9 mL of O2 (both measured at 315 K and 50.0 mm

Tanzania [10]

Answer:

a. Oxygen is the limiting reagent. $n_{SO_3}^{Theoretical}=8.096x10^{-4}mol SO_3$

b. $Y=58.9$ %

Explanation:

Hello,

a. Limiting reagent and sulfur trioxide's theoretical yield.

At first, we must compute the involved moles for both sulfur dioxide's and oxygen's as follows, considering the volumes in liters and the pressure in atm of 50.0mmHg*1atm/760mmHg=0.0658atm:

$n_{SO_2}=\frac{PV}{RT}=\frac{0.0658atm*0.2855L}{0.082\frac{atm*L}{mol*K}*315K} =7.273x10^{-3}molSO_2 \\n_{O_2}=\frac{PV}{RT}=\frac{0.0658atm*0.1589L}{0.082\frac{atm*L}{mol*K}*315K} =4.048x10^{-4}molO_2$

Afterwards, by considering the properly balanced chemical reaction:

$2SO_2(g)+O_2(g)-->2SO_3$

We compute the oxygen's moles that completely reacts with the previously computed $7.273x10^{-3}$ moles of $SO_2$ as follows:

$7.273x10^{-3}molSO_2*\frac{1molO_2}{2molSO_2} =3.6365x10^{-3}molO_2$

That result let us know that the oxygen is the limiting reagent since just $4.048x10^{-4}$ moles are available in comparison with the $3.6365x10^{-3}$ moles that completely would react with $7.273x10^{-3}$ moles of $SO_2$ .