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11111nata11111 [884]
3 years ago
9

Type the correct answer in the box. Express your answer to three significant figures. A balloon is filled with 0.250 mole of air

at 35°C. If the volume of the balloon is 6.23 liters, what is the absolute pressure of the air in the balloon? The absolute pressure of the air in the balloon is kilopascals.
Chemistry
1 answer:
Setler79 [48]3 years ago
7 0

Answer:

102.8 kPa.

Explanation:

  • We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm (P = ??? atm).  

V is the volume of the gas in L (V = 4000 mL = 6.23 L).  

n is the no. of moles of the gas in mol (n = 0.25 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),  

T is the temperature of the gas in K (T = 35ºC + 273 = 308 K).

∴ P = nRT/V = (0.25 mol)(0.0821 L.atm/mol.K)(308 K)/(6.23 L) = 1.015 atm.

  • To convert to kPa:

<em><u>multiply the pressure value by 101.325 </u></em>

∴ P = (1.0 atm)(101.325 kPa/1.0 atm) = 102.8 kPa.

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marin [14]

Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

Explanation:

1. Calculate the moles of copper(II) hydroxide

\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

6 0
3 years ago
Help please if its right i'll give brainiest
Mkey [24]
I believe it is spontaneous generation
5 0
3 years ago
A 232-lb fullback runs the 40-yd dash at a speed of 19.8 ± 0.1 mi/h.
Neporo4naja [7]

Answer:

(a)  7.11 x 10⁻³⁷ m

(b)  1.11 x 10⁻³⁵ m

Explanation:

(a)  The de Broglie wavelength is given by the expression:

λ = h/p = h/mv

where h is plancks constant, p is momentum which is equal to mass times velocity.

We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.

v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s

m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg

λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m

(b) For this part we have to use the uncertainty principle associated with wave-matter:

ΔpΔx > = h/4π

mΔvΔx > = h/4π

Δx = h/ (4π m Δv )

Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.

Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )  

     = 0.045 m/s

Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )

     = 1.11 x 10⁻³⁵ m

This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.

5 0
3 years ago
what is the molecular formula for a compound with the empirical formula: K2SO4 and a molecular mass of 696g​
LekaFEV [45]

Answer:

K8S4O16 or K8(SO4)4 depending on if the SO4 is supposed to represent sulfate or not

Explanation:

Find the molar mass of K2SO4 first:

2K + S + 4O ≈ 174 g/mol

Divide the goal molar mass of 696 by the molar mass of the empirical formula:

696 / 174 = 4

This means you need to multiply everything in the empirical formula by 4:

K2SO4 --> K8S4O16 or K8(SO4)4 depending on if the SO4 is for sulfate or not

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2 years ago
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Nata [24]

Explanation:

IM PRETTY SURE IT IS D !! IF ITS WRONG IM SORRY THAT WHAT

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