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3 years ago

list the following from greatest to least mass: 1 mol of H2, 2 moles of K, 3 moles of Mg, and 4 moles of He

1 answer:

3 0

H₂ = 2 g/mol

1 x 2 = 2 g of H₂
________________________

K = 39.09 g/mol

2 x 39.09 = 78.18 g of K
________________________________

Mg = 24.3 g/mol

3 x 24.3 = 72.9 g of Mg
_________________________________

He = 4 g/mol

4 x 4 = 16 g of He
___________________________________

Order : K , Mg , He and H₂

hope this helps!

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The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a

mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

3 0

3 years ago

How are the different colors generated by the flames? A. proton transitions B. reactions between different atoms C. electronic t

attashe74 [19]

Answer: C

Explanation:

According to Neils Bohr, atoms contain electrons which are arranged in energy levels. The energy levels proceed from the lowest to the highest. When energy is supplied to an atom,it moves from lower to higher energy levels. The higher energy level is known as the excited state. Excited states are short lived and atoms quickly return to ground state with emission of the absorbed energy in the form of visible light. This visible light must have one of the seven colours observed in the visible spectrum; Red, orange, yellow, indigo, blue, green, violet. Energy required for this excitation is supplied by heating the substance in a flame.

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3 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

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4 years ago

Answer (example of a bar graph)

Vilka [71]

Here's an example of a bar graph.

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3 years ago

.Which best describes the structure of 2-butene?

yan [13]

Answer:

Explanation:

The 2 - butene is an alkene with the formula C4H8, which can be used as the polybutene monomer or co-polymer. It is a compressed liquefied gas colorless, highly flammable characteristic odor. Prolonged exposure to this gas can cause frostbite causing a lowering of consciousness. Its formula is C4H8 and it is made up of 4 carbon atoms.

The best structure that describes 2-butene is that of Option A.

8 0

3 years ago