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galina1969 [7]
3 years ago
6

(Use your notes and info from active readings to explain WHY/HOW the science backs up your procedure) (BELOW)

Chemistry
1 answer:
FromTheMoon [43]3 years ago
8 0

Explanation:

It provides an objective, standardized approach to conducting experiments and, in doing so, improves their results. By using a standardized approach in their investigations, scientists can feel confident that they will stick to the facts and limit the influence of personal, preconceived notions.

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On which of the following factors does the amount of energy absorbed by an endothermic reaction depend?
jolli1 [7]

Answer: D!! ( difference in the potential energy of the reactants and products )

Explanation:

i have the same test

4 0
3 years ago
Need help ASAP!<br><br> How many moles of sodium nitrate are in 0.25 L of 1.2 M NaNO3 solution?
Setler79 [48]

Answer:

\boxed {\boxed {\sf 0.3 \ mol \ NaNO_3}}

Explanation:

Molarity is a measure of concentration in moles per liter.

molarity= \frac{moles \ of \ solute}{liters \ of \ solution}

The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.

  • molarity= 1.2 mol NaNO₃/L
  • liters of solution=0.25 L
  • moles of solute =x

1.2 \ mol \ NaNO_3/L= \frac{x}{0.25 \ L}

We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.

0.25 \ L *1.2 \ mol \ NaNO_3/L=\frac{x}{0.25 \ L} *0.25 \ L

0.25 \ L *1.2 \ mol \ NaNO_3/L=x

The units of liters cancel, so we are left with the units moles of sodium nitrate.

0.25  *1.2 \ mol \ NaNO_3=x

0.3 \ mol \ NaNO_3=x

There are 0.3 moles of sodium nitrate.

3 0
2 years ago
How many moles of water h2o are present in 75.0 g h2o?
nikklg [1K]
4.17 moles. Good luck! :)
7 0
3 years ago
What mass of butane in grams is necessary to produce 1.5×103 kJ1.5×103 kJ of heat? What mass of CO2CO2 is produced? Assume the r
saul85 [17]

32.8 g of Butane is required and 99.3 g of CO₂ is produced

<u>Explanation:</u>

The above mentioned reaction can be written as,

C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g)     where ΔH (rxn)= -2658 kJ

It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.

That is

$\frac{1500}{2658}=0.564 \text { moles }    of butane reacted

Now this moles is converted into mass by multiplying it with its molar mass  = 0.564 mol × 58.122 g / mol

                     = 32.8 g of butane.

Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol

                                        = 99.3 g of CO₂

Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced

7 0
3 years ago
Which of these is a common feature for the lac and trp operon? A. Regulatory proteins bind to operator B. Regulatory proteins bi
Arisa [49]
 C. <span>Regulatory proteins bind to repressor

Both produces certain proteins to break down lactose as a food source.</span>
4 0
3 years ago
Read 2 more answers
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