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lorasvet [3.4K]
3 years ago
8

How many significant figures are in 400..

Chemistry
2 answers:
marin [14]3 years ago
8 0

Answer:

3 significant figures are in 400..

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

  1. Digits from 1 to 9 are always significant and have infinite number of significant figures.
  2. All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.
  3. All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.
  4. All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.
  5. All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.
  6. In a non decimal number like 30 , there are 1 significant figure but number in which decimal is written after zero then all the zeros will become significant.

Given = 400.

As we can see that decimal point is present  after 0 which makes the all the zeros of the number significant.

So, significant figures are in 400. is 3.

Montano1993 [528]3 years ago
7 0
Answer:

There are 3 significant figures in 400.

Explanation:

All non-zero digits are significant, therefore the 4 is significant. Zeros after the number are only significant if there is a decimal point in the number, therefore the two zeros are also significant.
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At 400 k, the equilibrium constant for the reaction is kc = 7.0. br2 (g) + cl2 (g) 2brcl (g) a closed vessel at 400 k is charged
jeka57 [31]
a) when Kc = concentration of products / concentration of reactants
  So according  to the reaction equation:
Br2(g) + Cl2(g) → 2BrCl(g)

∴ Kc =[BrCl] ^2 / [Br2][Cl2]

b) when q = [BrCl]^2 / [Br2][Cl2]
and we have [BrCl] = 3 m 
[Br2] = 1 m 
[Cl2] = 1 m
So by substitution:
q= 3^2 / 1*1 = 9 

- and we can see that q > Kc 
the reaction is not at equilibrium that means there are more products and the reaction shifts to the left to increase the reactants and decrease the products to achieve equilibrium.

C) by using ICE table:

              Br2(g) + Cl2(g) → 2BrCl (g)
initial       1               1               3
change  -X              -X            +X
Equ       (1-X)          (1-X)         (3+X)

when Kc = [Brcl]^2/[Cl2][Br2]
by substitution:
7 = (3+X)^2 / (1+X) (1+X)  by solving this equation for X
∴X = 0.215
so at equilibrium:
∴ [Br2] = [Cl2] = 1-0.215 = 0.785 m 
    [BrCl] = 3+0.215 = 3.215 m
4 0
3 years ago
How do ionic bonds form??
mihalych1998 [28]
It is formed when a metal element is chemically combined to a nonmetal. The metal element will form a positive ion and the nonmetal will form a negative ion. They will then combined to form a very strong bond with very strong electrostatic forces between the particles.
4 0
3 years ago
How many moles of calcium chloride would react with 5. 99 moles of aluminum oxide?
slega [8]

There are 17.97 moles of calcium chloride would react with 5. 99 moles of aluminum oxide .

The balanced chemical equation between  reaction between calcium chloride and aluminum oxide is given as,

3CaCl_{2} (aq)+Al_{2} O_{3} (s) → 3CaO_{2} (s)+2Al Cl_{3} (aq)

The molar ratio of above reaction is  3:1

It means 3 moles of calcium chloride is require to react one mole of aluminum oxide.

The number of moles of calcium chloride requires to react with  5. 99 moles of  aluminum oxide  = 3 × 5. 99 = 17.97 moles

The equation in which number of atoms of elements in reactant side is equal to the number of atoms of elements in product side is called balanced chemical equation .

learn more about calcium chloride

brainly.com/question/15296925

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6 0
2 years ago
For the galvanic (voltaic) cell Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s) (E°= 0.77 V at 25°C), what is [Fe2+] if [Mn2+] = 0.040 M and
avanturin [10]

Answer:

0.01836 M

Explanation:

Again the reaction equation is;

Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)

E°cell= 0.77 V

Ecell= 0.78 V

[Mn2+] = 0.040 M

[Fe2+] = the unknown

n=2

From Nernst's equation;

Ecell= E°cell- 0.0592/n log Q

0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]

0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]

0.01/ -0.0296= log [Fe2+] /[0.040]

-0.3378= log [Fe2+] /[0.040]

Antilog(-0.3378) = [Fe2+] /[0.040]

0.459= [Fe2+] /[0.040]

[Fe2+] = 0.459 × 0.040

[Fe2+] = 0.01836 M

7 0
3 years ago
I have to do this for homework please help :)
kramer

Answer:

1..... nucleus

2......electron cloud

3.......protons

4........Neutrons

5..........electron

6............electrons

7...............Isotopes

8.....,...........ions

9....................charge

6 0
3 years ago
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