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2 years ago

What will be the volume of a balloon at 98 kPa if its volume is 9.65 L at 235 kPa?

1 answer:

4 0

<span>Boyle's Law or Boyle-Mariotte's Law: Under constant temperature, the volume occupied by a given fixed mass of a gas is inversely proportional to its pressure.

Formula:
</span> $\frac{ P_{1} }{ P_{2} } = \frac{ V_{2} }{ V_{1} }$
Product of extremes equals product of means
$P_{1} * V_{1} = P_{2} * V_{2}$

Data:
$P_{1} = 235\:kPa$
$V_{1} = 9.65\:L$
$P_{2} = 98\:kPa$
$V_{2} = ?$

Solving:
$P_{1} * V_{1} = P_{2} * V_{2}$
$235*9.65 = 98*V_{2}$
$2267.75 = 98V_{2}$
$98V_{2} = 2267.75$
$V_{2} = \frac{2267.75}{235}$
$\boxed{\boxed{V_{2} \approx 23.14\:L}}\end{array}}\qquad\quad\checkmark$

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In water, bases form _____.

atroni [7]

A base produces hydroxide ion in water

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3 years ago

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What is the [h3o ] for a solution of seawater at 25°c that has a ph of 8.3?

Kipish [7]

PH of solution at 25ºC = 8.3

$[ H_3O^+] = 10 ^{-pH}$

${H_3O^+] = 10 ^{-8.3}$

$[H_3O^+] = 5.011*10^{-9} M$

hope this helps!

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2 years ago

A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was fo

klio [65]

Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

$\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M$

4 0

2 years ago

If the net force on a object is 375 N to the right, in what direction will the object move?

Viktor [21]

To the right
Because there is an unbalanced force in that direction
:)

8 0

3 years ago

A gas has a volume of 590 mL at temperature of -55.0 C. What volume will the gas occupy at 30.0 C show your work

DENIUS [597]

Data:
$V_{initial} = 590\:mL$
$T_{initial} = -55.0^0C$
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → $T_{initial} = 218.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 30.0^0C$
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 → $T_{final} = 303.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }$
Product of extremes equals product of means:
$218.0* V_{f} = 590*303.0$
$218.0 V_{f} = 178770$
$V_{f} = \frac{178770}{218.0}$
$\boxed{\boxed{V_{f} \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark$

7 0

2 years ago