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solong [7]
4 years ago
11

Liz puts a 1 kg weight and a 10 kg on identical sleds. She then applies a 10N force to each sled. Which does not explain why the

smaller weight accelerate faster? A) The larger mass spends more force overcoming inertia. B) The smaller weight exerts less frictional force to overcome. Eliminate C) The acceleration depends directly on the mass of the object. D) The motion of the sled relates to action of unbalanced forces.
Chemistry
2 answers:
Fynjy0 [20]4 years ago
8 0

C ❤️ The acceleration depends directly on the mass of the object

Soloha48 [4]4 years ago
7 0

(05.01)

Which statement best explains if the graph correctly represents the proportional relationship y = 0.5x? (4 points)

<span>Yes, the points shown on the line would be part of y = 0.5xYes, all proportions can be shown on a graph of this lineNo, the points shown would not be part of y = 0.5xNo, proportions cannot be represented on a graph</span>
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Which two polymers have the same linkages bonding the monomers together?
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3 years ago
A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L
Pavel [41]

Answer:

The pH does not decrease drastically because the HCl reacts with the <u>sodium azide (NaN₃)</u> present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l)  ⇄ N₃⁻(aq) + H₃O⁺(aq)          

The pH of this solution is:

pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq)  ⇄  HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles

Now, the number of moles of HN₃ is:

\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles

Then, the pH of the buffer solution after the addition of HCl is:

pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.    

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

6 0
3 years ago
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