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Paraphin [41]
2 years ago
8

Large quantities of tritium were released into the atmosphere in the 1950s and 1960s during atmospheric nuclear testing. This tr

itium fell back to earth in precipitation. Assume that the precipitation joins runoff that goes directly to an open watershed. Where would you expect to find the tritium next? A. In a closed watershed B. In the atmosphere C. In an iceberg D. In the ocean
Chemistry
2 answers:
Sergeu [11.5K]2 years ago
8 0
The ocean
hope I helped
SVETLANKA909090 [29]2 years ago
4 0

Answer:  D. In the ocean

A watershed can be defined as regions of land where all surface water drains into the same place. It is a land that is occupied with water drained from interconnected  rivers, and lakes.  These drainage streams may aggregate in the ocean.

According to the above information the open watershed system likely to dispense off tritium in the ocean because the open watershed water may flow and combine with other reservoirs of water like ocean.

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What does X represent in the formula for the compound XCl4?
qaws [65]

I believe the answer would be C

The the compound of interest i.e.  XCl4, since there are 4 Cl atoms bonded to X. This signifies that the valency of X is 4.

There atomic number of C is 6. It's electronic configuration is giving by 1s2 2s2 2p2. Thus, there are 4 electrons in valence shell of C. This signifies that valency of C is 4. Hence the compound present in present case is CCl4.

Hope this helps :)


3 0
2 years ago
2 M n O 2 + 4 K O H + O 2 + C l 2 → 2 K M n O 4 + 2 K C l + 2 H 2 O , there are 100.0 g of each reactant available. Which reacta
Sliva [168]

Answer:

The limiting reactant is KOH.

Explanation:

To find the limiting reactant we need to calculate the number of moles of each one:

\eta = \frac{m}{M}

<u>Where</u>:

η: is the number of moles

m: is the mass

M: is the molar mass

\eta_{MnO_{2}} = \frac{100.0 g}{86.9368 g/mol} = 1.15 moles  

\eta_{KOH} = \frac{100.0 g}{56.1056 g/mol} = 1.78 moles  

\eta_{O_{2}} = \frac{100.0 g}{31.998 g/mol} = 3.13 moles  

\eta_{Cl_{2}} = \frac{100.0 g}{70.9 g/mol} = 1.41 moles  

Now, we can find the limiting reactant using the stoichiometric relation between the reactants in the reaction:

\eta_{MnO_{2}} = \frac{\eta_{MnO_{2}}}{\eta_{KOH}}*\eta_{KOH} = \frac{2}{4}*1.78 moles = 0.89 moles

We have that between MnO₂ and KOH, the limiting reactant is KOH.

\eta_{O_{2}} = \frac{\eta_{O_{2}}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{1}{1}*1.41 moles = 1.41 moles

Similarly, we have that between O₂ and Cl₂, the limiting reactant is Cl₂.

Now, the limiting reactant between KOH and Cl₂ is:

\eta_{KOH} = \frac{\eta_{KOH}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{4}{1}*1.41 moles = 5.64 moles

Therefore, the limiting reactant is KOH.

I hope it helps you!

6 0
2 years ago
Read 2 more answers
Here is a sketch of a 2px orbital: This sketch is about 800pm wide. The coordinate ( x , y , and z ) axes are also shown. You ca
amid [387]
<span><u>PA to PB 100 pm to the left of the nucleus, along the -x axis.</u>
<u>100 pm below the nucleus along the -z axis.</u>

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PAPB 100 pm to the right of the nucleus, along the +x axis. 100 pm above the nucleus, along the +z axis. </span>
7 0
2 years ago
In what situation do we use a volumetric flask, conical flask, pipette and graduated cylinder? Explain your answer from accuracy
Ad libitum [116K]
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Pippettes are of two types, volumetric and graduated. Pippettes are used where high accuracy is required and volumetric pippettes come in as little as 1 ml. Pippettes are usually used in titrations.
Graduated cylinders come in a wide variety of sizes and their accuracy can be down to as much as 1 ml. They are used to contain liquids.
3 0
2 years ago
Determine the y component of reaction at a using scalar notation. express your answer to three significant figures and include t
blagie [28]
In order to compute the y-component of a vector, we simply use the formula:

Fy = F*sin(∅)
Where ∅ is the angle of the vector measured from the positive x-axis and F is the magnitude of the vector.

Similarly, the x-component is calculated by substituting sin(∅) with cos(∅)
6 0
2 years ago
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