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3 years ago

In this simulation what do you think the bread and cheese separately represent, the atom or the molecule?

Chemistry

1 answer:

Svetllana [295]3 years ago

8 0

Answer:

the atom

Explanation:

The bread and cheese separately represent the atom because both bread and cheese are different from one another and we cannot assume it as a molecule because molecules formed when the group of atoms combine together by making bonds with each other and we know that bread and cheese did not make bonds with each other, they only attached, so we called them atoms not molecule.

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A plane has a cruising speed of 100 miles per hour when there is no wind. At this speed, the plane flew 600 miles with the wind

alina1380 [7]

The solution for this problem is:
Let x = speed of wind

Speed of plane with the wind = x + 100

Speed of plane against the wind = 100 -x

We will be using the formula for distance which is (Rate)(Time), getting the formula for time would be distance/rate Time to travel 600 miles with the wind = Time to travel 400 miles against the wind 600/(x + 100) = 400/(100 - x)

400(x + 100) = 600(100 - x)

400x + 40000 = 60000 - 600x

1000x = 20000

x = 20000/1000

x = 20 mph

8 0

3 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

What pressure is required to compress 196.0 liters of air at 1.83 atmosphere into a cylinder whose volume is 26.0 liters

drek231 [11]

Answer:

P₂ = 13.79 atm

Explanation:

Given data:

Initial volume = 196.0 L

Initial pressure = 1.83 atm

Final volume = 26.0 L

Final pressure = ?

Solution:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

Now we will put the values in formula,

P₁V₁ = P₂V₂

1.83 atm × 196.0 L = P₂× 26.0 L

P₂ = 358.68 atm. L / 26.0 L

P₂ = 13.79 atm

5 0

3 years ago

Question 1

Nesterboy [21]

P waves are faster S wave are slower P waves can pass through both solids and liquids S waves bounce of solids

3 0

3 years ago

Read 2 more answers

Question 231.15 pts A 0.200 M solution of a weak monoprotic acid HA is found to have a pH of 3.00 at room temperature. What is t

MariettaO [177]

Answer:

The correct answer is: Ka= 5.0 x 10⁻⁶

Explanation:

The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:

HA ⇄ H⁺ + A⁻

t= 0 0.200 M 0 0

t -x x x

t= eq 0.200M -x x x

At equilibrium, we have the following ionization constant expression (Ka):

Ka= $\frac{ [H^{+}] [A^{-} ]}{ [HA]}$