What is group 18 of the periodic table called?

At a festival, spherical balloons with a radius of 280. Cm are to be inflated with hot air and released. The air at the festival

evablogger [386]

Answer:

8608.18 balloons

Explanation:

Hello! Let's solve this!

Data needed:

Enthalpy of propane formation: 103.85kJ / mol

Specific heat capacity of air: 1.009J · g ° C

Density of air at 100 ° C: 0.946kg / m3

Density of propane at 100 ° C: 1.440kg / m3

First we will calculate the propane heat (C3H8)

3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J

Then we can calculate the mass of the air with the heat formula

Q = mc delta T

m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =

m = 93566.96kg

We now calculate the volume of a balloon.

V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3

Now we calculate the mass of the balloon

mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg

The amount of balloons is

93566.96kg / 10.87kg = 8608.18 balloons

5 0

3 years ago

Read 2 more answers

This number tells you the number of each type of atom in a compound.

KonstantinChe [14]

Answer:

avogadro's constant

Explanation:

this is the fixed number of the atoms in the molecule of an element

avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

that is all gases with same temperature and pressure will always have equal numbers of molecules

7 0

3 years ago

HELP ASAP ILL GIVE BRAINLIEST!!

Dahasolnce [82]

Of course we would experience them.

3 0

2 years ago

Determine how many gmol, kmol, and lbmols there are in 50 kilograms of n-hexane.

Artist 52 [7]

Answer: 581 gmol

0.581 kmol

$1.28\times 10^{-3}lbmol$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{50\times 1000g}{86g/mol}=581mol$

1. The conversion for mol to gmol

1 mol = 1 gmol

581 mol= $\frac{1}{1}\times 581=581gmol$

2. The conversion for mol to kmol

1 mol = 0.001 kmol

581 mol= $\frac{0.001}{1}\times 581=0.581kmol$

3. The conversion for mol to lbmol

1 mol = $2.2\times 10^{-3}lbmol$

581 mol= $\frac{2.2\times 10^{-3}}{1}\times 581=1.28\times 10^{-3}lbmol$

3 0

3 years ago

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0

o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2) Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min 0 10 20 30 40 ∞

conc B/(mol/L) 0 0.089 0.153 0.200 0.230 0.312

For a first order reaction:

$K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})$

where :

K = proportionality constant or the rate constant for the specific reaction rate

t = time of reaction

$C_o$ = initial concentration at time t

$C _{\infty}$ = final concentration at time t

$C_t$ = concentration at time t

To start with the value of t when t = 10 mins

$K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})$

$K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})$

$K_1 =0.03358 \ min^{-1}$

$K_1 \simeq 0.034 \ min^{-1}$

When t = 20

$K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})$

$K_2= 0.05 \times \ In ( 1.9623)$

$K_2=0.03371 \ min^{-1}$

$K_2 \simeq 0.034 \ min^{-1}$

When t = 30

$K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})$

$K_3= 0.0333 \times \ In ( \dfrac{0.312}{0.112})$

$K_3= 0.0333 \times \ 1.0245$

$K_3 = 0.03412 \ min^{-1}$

$K_3 = 0.034 \ min^{-1}$

When t = 40

$K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})$

$K_4=0.025 \times \ In ( \dfrac{0.312}{0.082})$

$K_4=0.025 \times \ In ( 3.8048)$

$K_4=0.03340 \ min^{-1}$

We can see that at the different time rates, the rate constant of $k_1, k_2, k_3, and k_4$ all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0

3 years ago