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shutvik [7]
2 years ago
9

By the way...... what do those symbols for the elements have in common (for number 7-11): Pb, Au, Cu, Hg, Na ? *

Chemistry
2 answers:
Keith_Richards [23]2 years ago
8 0

Answer:

those have symbols for their Latin or Greek name

Explanation:

hope it helps

galina1969 [7]2 years ago
8 0

Answer:

Na number of 11

Explanation:

Pb no. 82

Au no. 79

Cu no. 29

Hg no. 80

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Calculate the molecular mass of Al2(SO4)3(Molecular mass of Al=27, S=32, O=16) Pls fast
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Al₂(SO₄)₃

= 2.Al+3.S+12.O

= 2.27 + 3.32+12.16

= 54+96+192

=342 g/mol

8 0
1 year ago
Whats the summ of Mechanical energy
seropon [69]
the result of its motion or stored energy
4 0
2 years ago
A student made a sketch of a potential energy diagram to represent an exothermic reaction.
Montano1993 [528]

Answer:

Since heat is released for

C

3

H

8

(

g

)

+

5

O

2

(

g

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→

3

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2

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+

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2

O

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g

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+

2219.9

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,

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Δ

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−

2219.9 kJ/mol propane

. We approximate that this is the change in potential energy for the reactants going to the products.

https://www.khanacademy.org/

https://www.khanacademy.org/

The above is for an endothermic reaction.

A certain feature of combustion reactions suggests that you should draw the diagram DIFFERENTLY from what you see above. How should the energy of the products compare with that of the reactants for combustion reactions?

3 0
2 years ago
If it takes three "breathes" to blow up a balloon to 1.2 L, and each breath supplies the balloon with 0.060 moles of exhaled air
padilas [110]
A 3.0 L balloon is 2.5 times as big as a 1.2L balloon, and 3 times .060 equals .18, so if you multiply .18 by 2.5 you get .45 moles of air i’m pretty sure
3 0
3 years ago
If 5.00 grams of copper metal reacts with oxygen to produce 5.62 grams of 'copper oxide' ,what is the empirical formula of the '
a_sh-v [17]

Answer:

Cu_2O

Explanation:

Hello!

In this case, according to the law of conservation of mass, it is possible to realize that the mass of oxygen comes from the subtraction of the total mass and the mass of copper:

m_O=5.62g-5.00g\\\\m_O=0.62gO

Next, we compute the moles of both Cu and O given their atomic masses:

n_{Cu}=5.00gCu*\frac{1molCu}{63.546gCu}=0.0787molCu\\\\n_O=0.62gO*\frac{1molO}{16.0 gO}=0.03875molO

Now, we divide by the moles of oxygen as the fewest ones, in order to calculate their subscripts in the empirical formula:

Cu=\frac{0.0787}{0.03875} =2.0\\\\O=\frac{0.03875}{0.03875} =1

Therefore, the empirical formula turns out to be:

Cu_2O

Best regards!

3 0
2 years ago
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