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3 years ago

What amount (moles) of compound is present in 1.00 g of each of the compounds in Exercise 54?

Chemistry

1 answer:

Anika [276]3 years ago

5 0

Answer:

1. Caffeine, C₈H₁₀N₄O₂

Amount = 1.00/194 = 0.00515 moles

2. Ethanol, C₂H₅OH

Amount = 0.0217 moles

3. Dry Ice, CO₂

amount = 0.0227 moles

Note: The question is incomplete. The compound are as follows:

1. Caffeine, C₈H₁₀N₄O₂;

2. Ethanol, C₂H₅OH;

3. Dry Ice, CO₂

Explanation:

Amount (moles) = mass in grams /molar mass in grams per mole

1. Caffeine, C₈H₁₀N₄O₂

molar mass of caffeine = 194 g/mol

Amount = 1.00 g/194 g/mol = 0.00515 moles

2. Ethanol, C₂H₅OH

molar mass of ethanol = 46 g/mol

Amount = 1.00 g/46 g/mol = 0.0217 moles

3. Dry Ice, CO₂

molar mass of dry ice = 44 g/mol

amount = 1.00 g/44 g/mol = 0.0227 moles

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3 years ago

When a barium atom loses two electrons what charge does it have

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Answer:

It becomes a positive ion and its radius decreases

Explanation:

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3 years ago

A synthesis reaction takes place when carbon monoxide (CO) and hydrogen gas (H2) react to form methanol (CH3OH). How many grams

kakasveta [241]

Answer:

A. 3.2grams

Explanation:

First we need to get the chemical equation for this reaction:

CO + H₂ → CH₃OH

We then need to balance the equation:

CO + 2H₂ → CH₃OH

Our next step is to first convert our given into moles. We do this by figuring out first how many grams of each substance there are in 1 mole by adding up the atomic mass of each element in each substance.

Carbon(1) Oxygen(1)

CO = 12.011(1) + 15.999(1) = 28.01 g/mole

Hydrogen(2)

H₂ = 1.008(2) = 2.016g/mole

We can then use this to determine how many moles of each reactant we have given the mass.

$CO = 2.8g\\2.8g\times\dfrac{1mole}{28.01g}=0.1moles\\\\H_{2}=0.50g\\\\0.50g\times\dfrac{1mole}{2.016g}=0.469moles$

So here we have our new given:

0.1 moles of CO

0.496 moles of H₂

We then need to determine how much product we produce with our given.

According to our chemical equation we can assume that:

For every 1 mole of CO we can produce 1 mole of CH₃OH

Fore every 2 moles of H₂ we can produce 1 mole of CH₃OH

Using this ratio we can determine how much product each reactant will produce by using the ratios:

$0.1 moles of CO\times\dfrac{1moleofCH_{3}OH}{1moleofCO}=0.1moles of CH_{3}OH\\\\ 0.496molesofH_{2}\times\dfrac{1moleofCH_{3}OH}{2molesofH_{2}}=0.248molesofCH_{3}OH$

Now notice that they are not equal. The reactant that produces the least amount of product will be our limiting reactant. The limiting reactant determines the amount of product that is produced, because once it is completely used up, the reaction stops. So in this case, the amount of CH₃OH that is produced is 0.10 moles.

Now we need to figure out how many this is in grams. To do that we need to find out how many grams of CH₃OH there are in 1 mole of CH₃OH.

Carbon (1) Hydrogen(4) Oxygen(1)

CH₃OH = 12.001(1) + 1.008(4) + 15.999(1)

= 12.001 + 4.032 + 15.999 = 32.032g/mole

We then use this for converting:

$0.10 moles\times\dfrac{32.032g}{1mole}=3.2032g$

So the reaction will produce 3.2032g of CH₃OH or 3.2g of CH₃OH

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3 years ago

In what way would readings from a digital thermometer be preferable to those from a liquid-based thermometer

charle [14.2K]

A digital thermometer has less errors in reading temperatures because it displays the actual reading rather than approximating it in a liquid-based thermometer. Also, digital thermometer are advantageous to use in extreme temperature conditions because they can withstand in these type of conditions.

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3 years ago

Read 2 more answers

Tap water at room temperature has a pH of 7.2 and carbonate alkalinity of 200 mg/L (= 40 mmol/L). What is the concentration of b

UNO [17]

Answer:

the concentration of bicarbonate is [HCO₃⁻] = 0,03996 M and carbonate is [CO₃²⁻] = 3,56x10⁻⁵ M.

Explanation:

Carbonate-bicarbonate is:

HCO₃⁻ ⇄ CO₃²⁻ + H⁺ With pka = 10,25

Using Henderson-Hasselbalach formula:

pH = pka + log₁₀ $\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}$

7,2 = 10,25 + log₁₀ $\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}$

8,91x10⁻⁴ = $\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}$ (1)

Also:

0,040 M = [CO₃²⁻] + [HCO₃⁻] (2)

Replacing (2) in 1:

[HCO₃⁻] = 0,03996 M

Thus:

[CO₃²⁻] = 3,56x10⁻⁵ M

I hope it helps.

4 0

3 years ago