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olya-2409 [2.1K]
3 years ago
6

Placer deposits form when ____.

Chemistry
2 answers:
Oxana [17]3 years ago
8 0
Think it’s B or C because they make more sense
maria [59]3 years ago
8 0
The answer is B) <span>heavy eroded particles settle out of moving water</span>
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How many sulfur atoms are present in 100 grams of this compound? Report your answer to three significant figures.
kap26 [50]

Answer:

1.88 × 10²⁴ atoms

Explanation:

Step 1: Given data

Mass of sulfur: 100 g

Step 2: Calculate the moles corresponding to 100 g of sulfur

The molar mass of sulfur is 32.07 g/mol. The moles corresponding to 100 g of sulfur are:

100 g × (1 mol/32.07 g) = 3.12 mol

Step 3: Calculate the number of atoms in 3.12 moles of sulfur

We will use Avogadro's number: there are 6.02 × 10²³ atoms of sulfur in 1 mole of sulfur.

3.12 mol × (6.02 × 10²³ atoms/1 mol) = 1.88 × 10²⁴ atoms

7 0
4 years ago
What is the percent mass oxygen in calcium carbonate (CaCo3)?
Gnoma [55]
  <span>Step 1 is to determine the mass of each part 
Mass of Ca is 40.08 g 
Mass of C is 12.01 g 
Mass of O is 16.00 x 3 = 48.00 g 
Step 2 is to determine the total mass of the compound 
Total mass of CaCO3 is 40.08 + 12.01 + 48.00 = 100.09 g 

Step 3 is to determine the % of each part using the following formula: 
Mass of part / total mass x 100 = 

40.08 / 100.09 x 100 = 40.04 % Ca 

12.01 / 100.09 x 100 = 12.00 % C 

48.00 / 100.09 x 100 = 47.96 % O 

Step 4 is to double check by adding all percentages. If they equal 100, then I probably did it right. :) 
40.04 
+12.00 
+47.96 
=100.00</span><span>
</span>
4 0
3 years ago
Read 2 more answers
A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

3 0
3 years ago
Identify the substances that will appear in the equilibrium constant expression for the equation: 2Ag+(aq)+Zn(s)&lt;-&gt;Zn2+(aq
Simora [160]

Hey there!


We Know that:



 2 Ag⁺(aq) + Zn(s) <-> Zn²⁺(aq)+2 Ag(s)


The equilibrium expression for the reaction is:



Kc =  [ Zn⁺² ]  /  [Ag⁺ ]²


Hope that helps!

8 0
3 years ago
Read 2 more answers
How is data not actually obtained from the experiment represented in a line graph?
zavuch27 [327]
A colored line, as long as it is one single piece, not broken
5 0
3 years ago
Read 2 more answers
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