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The density of helium is 0.17 g/L. What Is the mass of helium in a 5.4 L helium balloon? (2 sig figs) *

ioda

Answer:

The correct answer is 0.92 g

Explanation:

The density is defined as the mass per unit of volume:

Density= mass/volume

From the data provided:

volume= 5.4 L

density= 0.17 g/L

Thus, to calculate the mass of helium:

mass= density x volume = 0.17 g/L x 5.4 L= 0.918 g ≅ 0.92 g

6 0

3 years ago

Why does the s orbital fill before the p orbitals?

worty [1.4K]

The answer is B. The s orbital holds a maximum of 2 electrons, where the p orbital holds a maximum of 6 electrons.

3 0

3 years ago

Read 2 more answers

A 4.98g sample of aniline (C6H5NH2, molar mass is 93.13g/mol) was combusted in a bomb calorimeter with a heat capacity of 4.25kJ

natima [27]

the value of delta H° for aniline = -3201. 5 Kj/mol

calculation

Step 1: find heat

Q ( heat) = C ( specific heat capacity) x ΔT ( change in temperature)

C= 4.25 kj/c°

ΔT = 69.8-29.5 = 40.3 c°

Q= 4.25 kj/c x 40.3 c = 171.28 kj

Step 2: find the moles of aniline

moles = mass/ molar mass

= 4.98 g/ 93.13 g/mol =0.0535 moles

Step 3 : find delta H

171.28 kj / 0.0535=3201.5 kj/mol

since the reaction is exothermic delta H = -3201.5 Kj/mol

3 0

3 years ago

What are the coefficients of the balanced reaction? _CH4 + _CI2 ➡ _CCI4 + _HCI

stealth61 [152]

Answer:

CH4 + 4Cl2 --> CCl4 + 4HCl

Explanation: Message me if you need a thorough explanation

7 0

3 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

Anna [14]

Answer: The number of moles of ethanol after equilibrium is reached the second time is 11. moles.

Explanation:

We are given:

Initial moles of ethene = 34 moles

Initial moles of water vapor = 15 moles

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 34 15

At eqllm: 34-x 15-x x

We are given:

Equilibrium moles of ethene = 24 moles

Equilibrium moles of water vapor = 5 moles

Calculating for 'x'. we get:

$34-x=24\\\\x=10$

Volume of container = 100.0 L

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CH_3CH_2OH]}{[CH_2=CH_2][H_2O]}$ .......(1)

$[CH_3CH_2OH]=\frac{10}{100}=0.1M$

$[CH_2=CH_2]=\frac{24}{100}=0.24M$

$[H_2O]=\frac{5}{100}=0.05M$

Putting values in expression 1, we get:

$K_c=\frac{0.1}{0.24\times 0.05}\\\\K_c=8.3$

Now, 11 moles of ethene gas is again added and equilibrium is re-established, we get:

The chemical equation for the formation of ethanol follows:

$CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH$

Initial: 24+11 5 10

At eqllm: 35-x 5-x 10+x

$[CH_3CH_2OH]=\frac{10+x}{100}$

$[CH_2=CH_2]=\frac{35-x}{100}$

$[H_2O]=\frac{5-x}{100}$

Putting values of in expression 1, we get:

$8.3=\frac{\frac{(10+x)}{100}}{\frac{(35-x)}{100}\times \frac{(5-x)}{100}}\\\\8.3=\frac{(10+x)\times 100}{(35-x)\times (5-x)}\\\\x^2-52x+55=0\\\\x=50.9,1.1$

The value of 'x' cannot exceed '35', so the numerical value of x = 50.9 is neglected.

Moles of ethanol = $(10+x)=10+1.1=11.$

Hence, the number of moles of ethanol after equilibrium is reached the second time is 11. moles.

8 0

3 years ago