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mash [69]
3 years ago
11

Which one is it ASAP

Chemistry
2 answers:
Gelneren [198K]3 years ago
6 0
4 is D. 5 is C. 6 is D i guess. 7 is D.
crimeas [40]3 years ago
4 0
Question 4 answer: Gravity only pulls large objects together. 

Question 5 answer: Friction

Question 6 answer: Unbalanced Force

Question 7 answer: Reaction Force.
Hope this helps!

Aye sir!
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How does the compound shown in part 5a of the transparency differ from the elements that comprise it?
Tresset [83]
<span>A compound is ''composed'' of elements. The periodic table is made up of elements. Atoms makes up elements and elements when reacted together make compounds. Na+ and Cl- makes NACL....salt. a compound</span>
5 0
3 years ago
Starting with the stock solution of 6.0 M, how many milliliters of 6.0 M sulfuric acid are needed to make 450 mL of 1.2 M soluti
Lina20 [59]

<u>Answer:</u> The volume of stock solution needed is 90 mL

<u>Explanation:</u>

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the stock sulfuric acid solution

M_2\text{ and }V_2 are the molarity and volume of diluted sulfuric acid solution

We are given:

M_1=6.0M\\V_1=?mL\\M_2=1.2M\\V_2=450mL

Putting values in above equation, we get:

6.0\times V_1=1.2\times 450\\\\V_1=\frac{1.2\times 450}{6.0}=90mL

Hence, the volume of stock solution needed is 90 mL

6 0
3 years ago
How many particles are in a 151 g sample of Li2O?
neonofarm [45]

Answer:

3.052 × 10^24 particles

Explanation:

To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)

The mass of Li2O given in this question is as follows: 151grams.

To convert this mass value to moles, we use;

moles = mass/molar mass

Molar mass of Li2O = 6.9(2) + 16

= 13.8 + 16

= 29.8g/mol

Mole = 151/29.8g

mole = 5.07moles

number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23

= 30.52 × 10^23

= 3.052 × 10^24 particles.

4 0
3 years ago
Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir
guajiro [1.7K]

Answer:

y_{O2} =4.3%

Explanation:

The ethanol combustion reaction is:

C_{2}H_{5} OH+3O_{2}→2CO_{2}+3H_{2}O

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}

Dividing the previous equation by x:

1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles

Calculate the number of moles of CO2 and water considering the same:

0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)

0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)

The total number of moles at the reactor output would be:

N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)

So, the oxygen mole fraction would be:

y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3%

6 0
3 years ago
why is it harder to remove an electron from fluorine than from carbon? to put it another way, why are the outermost electrons of
Verizon [17]

It is harder to remove an electron from fluorine than from carbon because the size of the nuclear charge in fluorine is larger than that of carbon.

The energy required to remove an electron from an atom is called ionization energy.

The ionization energy largely depends on the size of the nuclear charge. The larger the size of the nuclear charge, the higher the ionization energy because it will be more difficult to remove an electron from the atom owing to increased electrostatic attraction between the nucleus and orbital electrons.

Since fluorine has a higher size of the nuclear charge than carbon. More energy is required to remove an electron from fluorine than from carbon leading to the observation that;  it is harder to remove an electron from fluorine than from carbon.

Learn more: brainly.com/question/16243729

6 0
3 years ago
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