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Morgarella [4.7K]
3 years ago
15

PLEASE HELP ASAP WITH THESE THREE!

Chemistry
1 answer:
Furkat [3]3 years ago
5 0

Answer:

001: 71.2 g per ML

002: Consider a block of marble that occupies 287 cm3 and has a mass of 869 g.

a. What is its density?

d=m/v=869g/287 cm3=3.03 g/cm3

b. Will it float in:

H2O(l)? Why or why not?

Nope! Density > 1.00 g/mL

Hg(l) (density=13.55 g/mL)? Why or why not?

Yep! Density < 13.55 g/mL

003:  

Use the given functions to set up and simplify  

0.309 cm3

g/cm3=g/cm3

0.309cm3=g/cm3

 

Explanation:

http://childschemistry.weebly.com/uploads/7/8/6/3/78638416/density_key.pdf

Hope this helps I tried my best

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Aspirin, like many pharmaceutical drugs, can access the cell because it is a weak acid. This occurs because Choose one: A. it is
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Answer:

Correct answer is (D). as a weak acid it can cross the membrane when in its uncharged form.

Explanation:

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4 years ago
What is the empirical formula for a compound that is 29.44\% calcium, 23.55% sulfur, and 47.01% oxygen? This compound is a commo
PSYCHO15rus [73]

Answer:

Empirical formula is CaSO₄.

Explanation:

Given data:

Percentage of calcium =29.44%

Percentage of sulfur = 23.55%

Percentage of oxygen = 47.01%

Empirical formula = ?

Solution:

Number of gram atoms of Ca = 29.44 / 40 = 0.74

Number of gram atoms of S = 23.55 / 32 = 0.74

Number of gram atoms of O = 47.01 / 16 = 3

Atomic ratio:

            Ca                      :        S                :         O

           0.74/0.74           :     0.74/0.74      :       3/0.74

               1                     :          1              :          4

Ca : S : O = 1 : 1 : 4

Empirical formula is CaSO₄.

3 0
3 years ago
A polar covalent bond will form between which two atoms?
Alex_Xolod [135]

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If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Covalent

                

                Between 0.4 and 1.7 then it is Polar Covalent 

            

                Greater than 1.7 then it is Ionic

 

For Be and F,

                    E.N of Fluorine          =   3.98

                    E.N of Beryllium        =   1.57

                                                   ________

                    E.N Difference                2.41          (Ionic Bond)


For H and Cl,

                    E.N of Chorine           =   3.16

                    E.N of Hydrogen        =   2.20

                                                   ________

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For Na and O,

                    E.N of Oxygen          =   3.44

                    E.N of Sodium          =   0.93

                                                       ________

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For F and F,

                    E.N of Fluorine          =   3.98

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Result:

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