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3 years ago

What is the best way to make sure graphs are easy to interpret

Chemistry

1 answer:

Neporo4naja [7]3 years ago

7 0

<span>Use descriptive axis labels and legends
</span>

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liquid octane(CH3)(CH2)6CH3) reacts with gaseous oxygen gas(O2) to produce gaseous carbon dioxide(CO2) and gaseous water(H2O). I

max2010maxim [7]

The percent yield of carbon dioxide will be 49.0 %.

<h3>Percent yield</h3>

First, let's look at the equation of the reaction:

$2C_8H_1_8 + 25O_2 -- > 16CO_2 + 18H_2O$

The mole ratio of octane to oxygen is 2:25.

Mole of 3.43 g octane = 3.43/114.23 = 0.03 mol

Mole of 19.1 g oxygen = 19.1/32 = 0.60 mol

Thus, octane is limiting.

Mole ratio of octane to carbon dioxide = 2:16.

Equivalent mole of carbon dioxide = 0.03 x 8 = 0.24 mol

Mass of 0.24 mol carbon dioxide = 0.24 x 44.01 = 10.5624 grams

Percent yield of carbon dioxide = 5.18/10.5624 = 49.0 %

More on percent yield can be found here: brainly.com/question/17042787

#SPJ1

6 0

1 year ago

How many grams of CO2 could be formed with 2.09 x 1023 atoms of O?

choli [55]

Answer:

r u in high school this is hard

Explanation:

6 0

3 years ago

I need three examples of objects with high density and three with low density!!

Anuta_ua [19.1K]

Answer:

For example, a suitcase jam-packed with clothes and souvenirs has a high density, while the same suitcase containing two pairs of underwear has low density. Size-wise, both suitcases look the same, but their density depends on the relationship between their mass and volume. Mass is the amount of matter in an object.

5 0

3 years ago

Read 2 more answers

The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0

3 years ago

Describe the differences between polyp and medusa

stealth61 [152]

One difference is that some animals are polyp and some are medusa.

The other difference is that some animals have medusa in their life or polp in their life cycle.

Hope these two differences helps :D

5 0

2 years ago