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jarptica [38.1K]
3 years ago
7

Please someone help me with this!!

Chemistry
2 answers:
Mandarinka [93]3 years ago
6 0
You’re answer would be A, when you open a parachute it captures air and shoots you upwards for a couple seconds because of how fast it slows the weight down.
defon3 years ago
4 0

Answer:

agreed

Explanation:

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Arrow on the table below draw an
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What volume of methane gas at 237 K and 101.33 kPa do you have when the volume is decreased to 0.50 L, with a temperature of 300
Alexxandr [17]

Answer: A volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

Explanation:

Given: T_{1} = 237 K,   P_{1} = 101.33 kPa,      V_{1} = ?

T_{2} = 300 K,      P_{2} = 151.99 kPa,        V_{2} = 0.50 L

Formula used is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{101.33 kPa \times V_{1}}{237 K} = \frac{151.99 kPa \times 0.50 L}{300 K}\\V_{1} = 0.592 L

Thus, we can conclude that a volume of 0.592 L of methane gas is required at 237 K and 101.33 kPa when the volume is decreased to 0.50 L, with a temperature of 300 K and a pressure of 151.99 kPa.

8 0
3 years ago
Why do mentos react to coke? Also is this a chemical or physical reaction?
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Read 2 more answers
A chemistry student weighs out of lactic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to
34kurt

Answer:

28.0mL of the 0.0500M NaOH solution

Explanation:

<em>0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.</em>

<em />

The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:

H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O

<em>Where 1 mole of the acid reacts per mole of the base.</em>

<em />

You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.

the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:

0.126g ₓ (1mol / 90.08g) = <em>1.40x10⁻³ moles of H₃C-CH(OH)-COOH</em>

To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:

1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =

<h3>28.0mL of the 0.0500M NaOH solution</h3>
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Explain how Newton’s cradle demonstrates the law of conservation of energy.
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When power goes up it must come down.
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