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Talja [164]
2 years ago
7

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

Chemistry
1 answer:
Anika [276]2 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

<em>Note: The question is incomplete. The compound are as follows:</em>

<em> 1. Caffeine, C₈H₁₀N₄O₂;</em>

<em>2. Ethanol, C₂H₅OH;</em>

<em>3. Dry Ice, CO₂</em>

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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Mason notices that his boat sinks lower in the water in a freshwater lake than in the ocean. what could explain this
tino4ka555 [31]

Answer:

Mason notices that his boat sinks lower into the water in a freshwater lake than in the ocean. What could explain this?

Explanation:

The freshwater has less density then the ocean!

6 0
3 years ago
Some oxides are given below. (i)Na2O (ii)NO2 (iii) CO2 (iv) MgO a) Which are the basic oxides among these? b) What is the name o
umka2103 [35]
<h3>Further explanation</h3>

Basic oxides ⇒ metal(usually alkali/alkaline earth) +O₂

L + O₂ ⇒ L₂O

L + O₂ ⇒ LO

Dissolve in water becomes = basic solution

L₂O+H₂O⇒ 2LOH

LO + H₂O⇒ L(OH)₂

So the basic oxides : Na₂O and MgO

Na₂O + H₂O⇒NaOH

MgO +H₂O⇒Mg(OH)₂

The aqueous solution of CO₂(dissolve in water)

CO₂ + +H₂O⇒ H₂CO₃(carbonic acid)

5 0
3 years ago
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for
inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

Cu^{2+}(aq)+Fe(s)\rightarrow Cu(s)+Fe^{2+}(aq)

Reduction at cathode :

Cu^+(aq)+2e^-\rightarrow Cu(s)

Reduction potential of  Cu^{2+} to Cu=E^o_{1}=0.34 V

Oxidation at anode:

Fe(s)\rightarrow Fe^{2+}(aq)+2e^-

Reduction potential of  Fe^{2+} to Fe=E^o_{2}=-0.44 V

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}

Putting values in above equation, we get:

E^o_{cell}=0.34V -(-0.44 V)=0.78 V

The standard cell potential of the reaction is 0.78 Volts.

3 0
3 years ago
How many molecules of SrCrO4 are in a sample of SrCrO4 0.556 moles?
Amiraneli [1.4K]

Answer:

3.35*10^{23}\ SrCrO_4\ molecules

Explanation:

We\ are\ given\ that,\\No.\ of\ moles\ of\ SrCrO_4=0.556\\Hence,\\As\ we\ know\ that,\\No.\ of\ particles=Avagadro's\ Constant*No.\ of\ moles\\We\ already\ know\ that\ Avagadro's\ Constant=6.022*10^{23}\\Here,\\No.\ of\ SrCrO_4\  molecules= 6.022*10^{23}*0.556\\Hence,\\No.\ of\ SrCrO_4\  molecules=3.348*10^{23} molecules\ \approx 3.35*10^{23}\ SrCrO_4\ molecules

7 0
2 years ago
If a 2.50 liter container is filled with Ne gas at a pressure of 650 mm Hg and at 25°C, what mass of Ne is in the container?
seraphim [82]

Answer:

1.76 g is the mass of Ne is in the container.

Explanation:

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 650 mm Hg

V = Volume of the gas = 2.50 L

T = Temperature of the gas = 25^oC=[25+273]K=298K

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

n = number of moles of Ne gas = ?

Putting values in above equation, we get:

650mmHg\times 2.50L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{650\times 2.50}{62.3637\times 298}=0.0874mol

Also, molar mass of Ne = 20.1797 g/mol

So, Mass = Moles\times Molar\ mass = 0.0874\times 20.1797 g = 1.76\ g

<u>1.76 g is the mass of Ne is in the container.</u>

7 0
3 years ago
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