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Fantom [35]
2 years ago
8

A piece of wire contains 1.52x10^22 atoms of copper. Calculate the miles of copper in the wire

Chemistry
1 answer:
melamori03 [73]2 years ago
6 0

Moles of Copper =

1.52×10^{22}atoms×\frac{1 mol }{6.022 * 10^{23}atoms}

= 0.0252 mol Cu

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A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration
Sliva [168]

Answer:

<h2>The equilibrium constant Kc for this reaction is 19.4760</h2>

Explanation:

The volume of vessel used= 100 ml

Initial moles of NO= \frac{2.60}{10^2} moles

Initial moles of H2= \frac{1.30}{10^2} moles

Concentration of NO at equilibrium= 0.161M

Concentration(in M)=\frac{moles}{volume(in litre)}

Moles of NO at equilibrium= 0.161(\frac{100}{1000})

                                            =\frac{1.61}{10^2} moles

               

                    2H2 (g)        +    2NO(g) <—>    2H2O (g) +    N2 (g)

<u>Initial</u>          :1.3*10^-2          2.6*10^-2                0                   0        moles

<u>Equilibrium</u>:1.3*10^-2 - x     2.6*10^-2-x              x                   x/2     moles

∴\frac{2.60}{10^2}-x=\frac{1.61}{10^2}

⇒x=\frac{0.99}{10^2}

Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)

<u>Equilibrium</u>:0.31*10^-2      1.61*10^-2          0.99*10^-2        0.495*10^-2  moles

⇒Kc=\frac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2}  (0.1)

⇒Kc=19.4760

3 0
2 years ago
Trimix is a general name for a type of gas blend used by technical divers and contains nitrogen, oxygen and helium. In one Trimi
gladu [14]

Answer:

The correct answer is 28.2 %.

Explanation:

Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,  

Ptotal = PO2 + PN2 + PHe

= 55 + 90 + 50

= 195 atm

The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100

Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,  

%VO2 = (55/195) * 100 = 28.2%

%VN2 = (90/195) * 100 = 46%

%VHe = (50/195) * 100 = 25.64%

Hence, the percent oxygen by volume present in the blend is 28.2 %.  

8 0
2 years ago
a student adds 15 g of baking soda to 10 g of acetic acid in a beaker. a chemical reaction occurs and a gas is given off. after
Zanzabum

The chemical equation representing the reaction between baking soda and acetic acid :

NaHCO_{3}(aq)+CH_{3}COOH(aq)-->CH_{3}COONa(aq)+H_{2}O(l)+CO_{2}(g)

According to the Law of conservation of mass, the overall mass is conserved in a chemical reaction. That is total mass of reactants is equal to the mass of the products.

Given here Mass of Baking soda = 15 g

Mass of acetic acid = 10 g

Total mass of the reactants = 15 g +10 g = 25 g

After the reaction, total mass of products = 23 g

This difference in masses must be because one of product carbon dioxide is a gas, which is lost and hence the mass of products is lower than the mass of reactants. Therefore, the mass is conserved in this reaction.

7 0
2 years ago
A food web showing the flow of energy through a freshwater ecosystem is shown below. Which of the animals shown in the food web
kati45 [8]

Answer: minnows

Explanation: tiny fish

7 0
2 years ago
Read 2 more answers
The terms motif (fold) and domain describe levels of protein organization more complicated than primary or secondary structure.
posledela

Answer:

Each specific property of motif and domain is explained.

Explanation:

Domain;

  • May retain a 3D structure when separated from rest of the protein.          
  • Unit of tertiary structure because alpha helix and beta sheets are units of secondary structure.
  • Stable globular units like pyruvate kinase
  • May be distinct functional units in a protein

Motif;

  • Repetetive supersecondary structure because they contain cluster of secondary structure.
  • Beta Alpha Beta unit is an example of motif
  • Clusters of secondary structure

Both Motif and Domain;

  • Stabilized by hydrophobic interactions like hydrogen bonding stabilize the both.
  • Depends on primary structure like the arrangement of amino acid in polypeptide chain determine the secondary and tertiary structure of proteins.

8 0
2 years ago
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