Answer:
B. control rods and moderators
Unlike solid matter, where particles are tightly packed and slightly vibrating, or gas, where particles go around everywhere and are extremely loose, a liquid has particles that are loosely packed but are still in slight contact with each other. Hope that's good enough
Answer:
We would need 10 mL of the concentrated CaCl₂ stock solution, and 30 mL of water.
Explanation:
To solve the question asked we can use the C₁V₁=C₂V₂ equation, where:
We <u>solve for V₁</u>:
We would need 10 mL of the concentrated CaCl₂ stock solution, and (40-10) 30 mL of water.
Answer:
the number of milliliters of a 1M is 402mL
Explanation:
The computation of the number of milliliters could be determined by using the following formula
As we know that

where,
V_1 and V_2 are the starting and final volumes
And, the M_1 and M_2 are the starting and the final molarities
Now the V_1 is

So, the V_1 is 402mL
Hence, the number of milliliters of a 1M is 402mL
The mass of sodium sulfite that was used will be 1,890 grams.
<h3>Stoichiometric problems</h3>
First, the equation of the reaction:

The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.
Mole of 960 grams SO2 = 960/64 = 15 moles
Equivalent mole of sodium sulfite that reacted = 15 moles
Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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