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LenKa [72]
3 years ago
5

Which of the following processes are physical changes?

Chemistry
2 answers:
lubasha [3.4K]3 years ago
8 0
C) cutting a snowflake out paper
because someone has to physically cut the paper into a snowflake
Stolb23 [73]3 years ago
7 0
C) and d) are physical changes while b) is chemical
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When sodium chloride reacts with iron it forms sodium and iron (ll) Chloride. Write down the balanced equation and classify it \
BartSMP [9]

Answer:

NaCl + Fe2+  ---> FeCl2 + Na+

Single replacement

Explanation:

Na is replaced by Fe2+ to form FeCl2, and that is the only replacement. Thus, it is single replacement.

7 0
3 years ago
A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
3 years ago
What is the number of protons plus the number of neutrons in the nucleus of an atom equivalent to?
Kryger [21]

Answer:

b

Explanation:

b/c proton + neutron=mass number and

mass number - proton= neutron

7 0
2 years ago
How do you convert between the mass and the number of moles of a substance?
Ksivusya [100]
To convert from grams to moles, you divide the number of grams by the molar mass To convert  from moles to grams, you multiply by the molar mass. You must first calculate the molar mass of the substance
6 0
3 years ago
The electron config for 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p3​
nadya68 [22]

Answer:

I don't understand what you are asking

3 0
3 years ago
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