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choli [55]
3 years ago
5

Wedges and dashes are used to indicate the ____________ shape in a drawn structure. A wedge indicates that the atom or group is

_________ of the page and a dash indicates that the atom or group is going into the page .
Chemistry
1 answer:
OverLord2011 [107]3 years ago
4 0

Answer:

Wedges and dashes are used to indicate the three dimensional shape in a drawn structure.

A wedge indicates that the atom or group is GOING OUT of the page.

Explanation:

wedges and dashes are ways or drawings use to represent three dimensional shape in a structure. Lines are use to represent image in the structure. A wedge line shows that molecules or bonds are moving towards the viewer i.e out of the the page.

A dashes line shows that bonds or atom are moving away the viewer or going into the group.

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If an atom has 30 electrons and 35 neutrons, what is its approximate atomic mass? what is the name of this element?
Murljashka [212]
30 electrons = atomic number
Look at your periodic table. You will see zinc has a atomic number (the top number) of 30.

When you add 30 and 35 you get 65. The is the atoms ATOMIC MASS. So looking at zinc on the periodic table, look at its bottom number, which is its atomic mass. It will be 65.

This means the name of the element is ZINC.

Hope this helped!
6 0
3 years ago
What type of weather will maryland expect in a day or so?
ycow [4]

Answer:

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Explanation:

8 0
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Choose from the following descriptions of solid:
mixas84 [53]
Pretty sure it's C!
3 0
3 years ago
Solve an equilibrium problem (using an ice table) to calculate the ph of each solution: part a a solution that is 0.195 m in hc2
IrinaK [193]
To determine the pH of a solution which has 0.195 M hc2h3o2 and 0.125 M kc2h3o2, we use the ICE table and the acid dissociation constant of hc2h3o2 <span>to determine the concentration of the hydrogen ion present at equilibrium. We do as follows:

HC2H3OO = H+ + </span>C2H3OO-
KC2H3OO = K+ + C2H3OO-

Therefore, the only source of hydrogen ion would be the acid. We use the ICE table,
                    HC2H3OO           H+        C2H3OO-
I                     0.195                  0              0.125
C                      -x                    +x               +x
------------------------------------------------------------------
E                 0.195-x                x              0.125 + x

Ka = <span>1.8*10^-5 = (0.125 + x) (x) / 0.195 -x 
x = 2.81x10^-5 M = [H+]

pH = - log [H+]
pH = -log 2.81x10^-5
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Therefore, the pH of the resulting solution would be 4.55.
</span>
4 0
3 years ago
Calculate the mass of glucose metabolized by a 60.0 −kg person in climbing a mountain with an elevation gain of 1550 m . Assume
lbvjy [14]

Answer:

Mass of glucose = 515.34 g

Explanation:

We are given;

Mass; m = 60 kg

Elevation; h = 1550 m

Acceleration due to gravity; 9.8 m/s²

Now, work performed to lift 60kg by 1550m is given by the formula;

W = mgh

W = 60 × 9.8 × 1550

W = 911400 J

We are told the actual work is 4 times the one above.

Thus;

Actual work = 4W = 4 × 911400 = 3,645,600 J

Now,

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We are given standard enthalpy of combustion = -1273.3 KJ/mol = -1273300

Moles of glucose = 3645600/1273300 = 2.863mol

Mass of glucose = 2.863 mol × 180 g/mol

Mass of glucose = 515.34 g

4 0
4 years ago
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