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saul85 [17]
3 years ago
14

The photo shows the San Andreas fault zone. This is a fault formed at a transform boundary.

Chemistry
2 answers:
brilliants [131]3 years ago
8 0

Answer:

The answers is Earthquakes

nexus9112 [7]3 years ago
7 0

Answer:

it is earthquakes

Explanation:

i just took it

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Predict whether the following reactions will be exothermic or endothermic.
Solnce55 [7]

Answer:

A. N₂(g) + 3H₂(g) -----> 2NH₃    exothermic

B. S(g) + O₂(g) --------> SO₂(g)    exothermic

C. 2H₂O(g) --------> 2H₂(g) + O₂(g) endothermic

D. 2F(g) ---------> F₂(g) exothermic

Explanation:

The question says predict not calculate. So you have to use your chemistry knowledge, experience and intuition.

A. N₂(g) + 3H₂(g) -----> 2NH₃    is exothermic because the Haber process gives out energy

B. S(g) + O₂(g) --------> SO₂(g)    is exothermic because it is a combustion. The majority, if not all, combustion give out energy.

C. 2H₂O(g) --------> 2H₂(g) + O₂(g) is endothermic because it is the reverse reaction of the combustion of hydrogen. If the reverse reaction is exothermic then the forward reaction is endothermic  

D. 2F(g) ---------> F₂(g) is exothermic because the backward reaction is endothermic. Atomisation is always an endothermic reaction so the forward reaction is exothermic

5 0
3 years ago
When a force is applied to an object, its _____, the object _________________________________________________________.
Arada [10]
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7 0
3 years ago
Determine the energy change in the following reaction. This reaction is considered ...
Juliette [100K]

Answer:isothermic

Explanation:

8 0
2 years ago
Read 2 more answers
What part of the microscope is the letter k?
snow_tiger [21]

Answer:

if you are asking k then the round one is condenser

if not then its a stage clip

6 0
2 years ago
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M 

CH3COONa = 0.0076 / 0.071 = 0.1070M 

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74. 

pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

pH = 4.74 + log ( 0.1070/0.1493) 

pH = 4.74 + log 0.717 

pH = 4.74 + (-0.14) 

pH = 4.60.

7 0
3 years ago
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