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mr Goodwill [35]
3 years ago
10

Do the following calculation and express the answer using correct scientific notation. (6.00x10^23)y(3.00)/284)

Chemistry
1 answer:
Naily [24]3 years ago
5 0
I know its not the answer but i had to put something bc im jus setting this up im rlly sorry for the inconvenience but i looked it up n got a quizlet n the answer on there was 1.81x10^-4 if it helps
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What would happen to the annual rates of limestone (CaCO3) precipitation in the ocean if the ocean water were to become much war
stiv31 [10]

Answer:

CaCO3 exoskeleton dissolves in acidic water

Explanation:

The increasing CO2 level makes the ocean water acidic and hence reduces the pH. In such acidic environment, marine organism that produce calcium carbonate shells or skeletons are negatively affected. Coral reefs and coralline algae abilities to produce skeleton also reduces.  

Calcium carbonate dissolves in acid. Thus, the more acidic the ocean water is the faster and easier it is to dissolve the exoskeleton and shell of marine organisms made up of calcium carbonate  

4 0
3 years ago
Please help ASAP! Will mark brainiest if correct
Scilla [17]

Answer: 1 is A

2 is C

3 is D

Hope this help

7 0
3 years ago
a solution of sodium hydroxide, 0.0500 M, is used to titrate a 15.00 mL sample of hydrochloric acid to the endpoint. The initial
Akimi4 [234]

Answer:

The answer to your question is Molarity = 0.0708

Explanation:

Data

NaOH  0.05 M     Volume 1 = 3.87 ml    Volume 2 = 25.11 ml

HCl   15 ml

Process

1.- Find the volume used of NaOH

                              25.11 - 3.87 = 21.24 ml = 0.02124 l

2.- Write the balanced equation of the reaction

                   NaOH  +  HCl   ⇒   NaCl + H₂O

3.- Calculate the moles of NaOH in the solution

Molarity = \frac{moles}{volume}

moles = Molarity x volume

moles = 0.05 x 0.02124

moles = 0.001062

4.- From the reaction we know that NaOH and HCl react in a proportion 1:1.

                   1 mol of NaOH -------------  1 mol of HCl

 0.001062 moles of NaOH ------------    x

                  x = (0.001062 x 1) / 1

                  x = 0.001062 moles of HCl

5.- Find the molarity of HCl

Molarity = \frac{0.001062}{0.015}

Molarity = 0.0708

4 0
3 years ago
Which of the following statements, if true, would support the claim that the NO3− ion, represented above, has three resonance st
tester [92]

Answer:

One of the bonds in nitrate is shorter than the other two.

Explanation:

We would firstly need to draw the Lewis structure for nitrate anion. To do this, let's follow the standard steps:

  • calculate the total number of valence electrons: five from nitrogen, each oxygen contributes 6, so a total of 18 from oxygen atoms, as well as one from the negative charge, we have a total of 24 valence electrons;
  • assign the central atom, usually this is the atom which is single; in this case, we have nitrogen as our central atom;
  • assign single bonds to all the terminal atoms (oxygen atoms);
  • assign octets to the terminal atoms and calculate the number of electrons assigned;
  • the number of electrons assigned is 24, so no lone pairs are present on nitrogen;
  • calculate the formal charges: each oxygen has a formal charge of -1 (formal charge is calculated subtracting the sum of lone pair electrons and bonds from the number of valence electrons of that atom); nitrogen has a formal charge of +2;
  • nitrogen doesn't have an octet as well, so we'll both minimize its formal charge and make it obtain an octet if we make one double bond N=O.

Therefore, we may have 3 resonance structures, as this double bond might be formed with any of the 3 oxygen atoms.

By definition, double bonds are shorter than single ones, so one of the bonds is shorter than the other two.

7 0
3 years ago
How much energy is required when a 142.1 gram sample of aluminum goes from 25.5°C to 46°C? (the specific heat of aluminum is 0.9
vfiekz [6]

Answer:

2621.75 j heat is required to increase the temperature 25.5°C to 46°C.

Explanation:

Given data:

Mass of sample = 142.1 g

Initial temperature = 25.5°C

Final temperature = 46°C

Specific heat capacity of Al = 0.90 J/g.°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 46°C - 25.5°C

ΔT = 20.5°C

Q = 142.1 × 0.90 J/g.°C × 20.5°C

Q = 2621.75 j

Thus,  2621.75 j heat is required to increase the temperature 25.5°C to 46°C.

4 0
3 years ago
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