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Oksana_A [137]
3 years ago
15

HELP ILL GIVE U FREE POINTS AND BRAINLEST !! Observe the plate boundary shown here. What type of plate boundary is this? Many ge

ologic events occur at these boundaries. Some are shown, but one is missing. Examine the diagram. Select the geologic event you would predict occurring at this plate boundary?
A) convergent oceanic-oceanic plate boundary; volcano
B) convergent oceanic-continental plate boundary; earthquake
C) divergent oceanic-oceanic plate boundary; sea floor spreading
D) convergent continental-continental plate boundary; mountain building

Chemistry
2 answers:
Svetlanka [38]3 years ago
6 0

Answer:

here, idk for sure but

Explanation: i think its c

Studentka2010 [4]3 years ago
3 0
The answer is B.
You can rule C out because divergent means moving away. Rule out A because there is an oceanic and continental plate, not 2 of the same type. Rule D out for the same reason.
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3 years ago
How many grams of hydrochloric acid will react completely with a block of gold is 3.2 cm by 3.8 cm by 2.8 cm, if the density of
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mass of HCl = 243.5426 grams

Explanation:
1- we will get the mass of the reacting gold:
volume of gold = length * width * height
volume of gold = 3.2 * 3.8 * 2.8 = 34.048 cm^3 = 34.048 ml<span>
density = mass / volume
Therefore:
mass = density * volume
mass of gold = </span>19.3 * 34.048 = 657.1264 grams

2- we will get the number of moles of the reacting gold:
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number of moles = 657.1264 / 196.96657 
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3- we will get the number of moles of the HCl:
First, we will balanced the given equation. The balanced equation will be as follows:
Au + 2HCl ......> AuCl2 + H2
This means that one mole of Au reacts with 2 moles of HCl.
Therefore 3.3362 moles will react with 2*3.3362 = 6.6724 moles of HCL

4- we will get the mass of the HCl:
From the periodic table:
molar mass of H = 1 gram
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Therefore:
molar mass of HCl = 1 + 35.5 = 36.5 grams/mole
number of moles = mass / molar mass
Therefore:
mass = number of moles * molar mass
mass of HCl = 6.6724 * 36.5
mass of HCl = 243.5426 grams

Hope this helps :)
4 0
3 years ago
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