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3 years ago

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How hot, in degrees Celsius, would the air inside the balloon have to get in order for the balloon to lift off the ground? Assum

e the molar mass of air is 28.97 g/mol and its density is 1.20 kg/m3.

1 answer:

SVETLANKA909090 [29]3 years ago

3 0

Answer:

The temperature in degress Celsius is 52.25°C

Explanation:

According the equation:

$V(\rho _{a} -\rho _{t})g=mg\\\rho _{a} -\rho _{t}=\frac{n}{V} =\frac{340}{2950} =0.115kg/m^{3}$

$\rho _{t}=\rho _{a} -0.115=1.2-0.115=1.085kg/m^{3}$

The temperature is:

$T=\frac{P}{\rho _{t} R} =\frac{1.013x10^{5} }{1.085*287.05} =325.25K=52.25$ °C

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Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent

sladkih [1.3K]

Answer:

$\large\boxed{\large\boxed{0.0014M/s}}$

Explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

<u>1. Table</u>

Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s)

1 0.20 0.20 0.20 6.0×10⁻⁵

2 0.20 0.20 0.60 1.8×10⁻⁴

3 0.40 0.20 0.20 2.4×10⁻⁴

4 0.40 0.40 0.20 2.4×10⁻⁴

<u>2. Orders</u>

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

0.60 / 0.2 = 3, and

1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

0.40/0.2 = 2

2.4×10⁻⁴ / 6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

<u>3. Rate equation</u>

Ther orders are:

$a=2\\\\b=0\\\\c=1$

Hence the rate is:

$rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C$

<u>4. Rate constant, k</u>

<u />

You can use any trial to find the value of the constant, k

Using trial 1:

$6.0\times 10^{-5}M/s=k(0.20M)^2(0.20M)\\ \\ k=\frac{6.0\times 10^{-5}M/s}{(0.20M)^2(0.20M)}=0.0075M^{-2}s^{-1}$

<u>5. Rate law:</u>

$rate=k[A]^2C=0.0075[A]^2[C]$

<u>6. Substitute</u>

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

$rate=0.0075M^{-2}s^{-1}[A]^2[C]=0.0075M^{-2}s^{-1}[0.45M]^2[0.9M]$

$r=0.00136688M/s\approx 0.0014M/s$

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3 years ago

PLZ HELP (SCIENCE) Match the following items.

FromTheMoon [43]

Oxidation: a chemical change resulting form a reaction with oxygen
Endothermic:a reaction that takes in energy
Exothermic: a reaction that releases energy
Decomposition:chemical change whereby a molecule breaks down into simpler moles clues or elements
Subscript:written underneath or below

6 0

2 years ago

Read 2 more answers

Use the following information S(s)+O2(g)→SO2(g), ΔH∘ = -296.8kJ SO2(g)+12O2(g)→SO3(g) , ΔH∘ = -98.9kJ to calculate ΔH∘f for H2SO

Irina-Kira [14]

Answer : The enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation of $H_2SO_4$ will be,

$S+H_2+2O_2\rightarrow H_2SO_4$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $S+O_2\rightarrow SO_2$ $\Delta H_1=-298.2$

(2) $SO_2+\frac{1}{2}O_2\rightarrow SO_3$ $\Delta H_2=-98.2$

(3) $SO_3+H_2O\rightarrow H_2SO_4$ $\Delta H_3=-130.2$

(4) $H_2+\frac{1}{2}O_2\rightarrow H_2O$ $\Delta H_4=-285.8$

Now adding all the equations, we get the expression for enthalpy of formation of $H_2SO_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-298.2)+(-98.2)+(-130.2)+(-285.8)$

$\Delta H=-812.4kJ/mol$

Therefore, the enthalpy of formation of $H_2SO_4$ is, -812.4 kJ/mole

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3 years ago

What minimum volume of 0.200 m potassium iodide solution is required to completely precipitate all of the lead in 195.0 ml of a

lbvjy [14]

First, we write the reaction equation:

2KI + PbNO₃ → K₂NO₃ + PbI₂
The molar ratio of KI to PbNO₃ is 2 : 1
Moles of PbNO₃ present:
Moles = concentration (M) x volume (dm³)
= 0.194 x 0.195
= 0.038
Moles of KI required = 2 x 0.038 = 0.076 moles
concentration = moles / volume
volume = moles / concentration
= 0.076 / 0.2
= 0.38 L = 380 ml

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3 years ago

Condensation point and freezing point of argon in KELVIN.

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Answer:

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