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GuDViN [60]
3 years ago
14

What is group 18 of the periodic table called?

Chemistry
2 answers:
Korolek [52]3 years ago
4 0
Group 18 is known as the Noble/ Inert Gases
Stolb23 [73]3 years ago
4 0

Group 18 is called the noble glass


Sorry If I am wrong


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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<em> </em><em>at</em><em> </em><em>the</em><em> </em><em>same</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>pressure</em><em> </em><em> </em><em>contain</em><em> </em><em>equal</em><em> </em><em>numbers</em><em> </em><em>of</em><em> </em><em>molecules</em><em> </em>

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

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
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