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PtichkaEL [24]
2 years ago
15

As Jesse's hybrid accelerates to pass a car on the highway, he notices that his gas mileage drops from 40 miles per gallon to 15

. One likely explanation for this is:
A. the engine is powered by a biofuel that is not needed while cruising.
B. his car operates on battery power when he begins to accelerate.
C. his car has turned on more pistons to provide the extra energy needed to accelerate.
D. his car operates only on battery power on the highway.
Chemistry
2 answers:
e-lub [12.9K]2 years ago
7 0
The most likely explanation for this observation is C. his car has turned on more pistons to provide the extra energy needed to accelerate.

When cruising, hybrid cars are able to employ electrical energy to drive the car. Moreover, even if a vehicle is not a hybrid, a greater amount of fuel is consumed when one accelerates because the vehicle has to generate a force larger than the force of air resistance in order for it to accelerate. This increased demand of force reduces the vehicle's fuel economy.
kolezko [41]2 years ago
3 0

Jesse's car is a hybrid that shuts off extra pistons when it is not accelerating APEX

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If you perform the experiment described in investigation #15 by mixing 10 g of glue with 13 g of water and 8 g of sodium borate
Phantasy [73]

10 g of glue with 13 g of water ,

Mass ratio of the material can be calculated as:    

\frac{Mass of glue}{Mass of glue + Mass of water}  

\frac{10 g}{10 g + 13 g }  

\frac{10 g}{23 g }

8 g of sodium borate suspended in 11 g of water, mass ratio can be calculated as:

\frac{Mass of sodium borate}{Mass of sodium borate + Mass of water}  

\frac{ 8 g}{ 8 g + 11 g }  

\frac{8 g}{19 g }

3 0
2 years ago
Calculate the mass of 25,000 molecules of nitrogen gas. (1 mole = 6.02 x 1023 molecules)
Ainat [17]

Hey there!

Molar mass N2 = 28.01 g/mol

Therefore:

28.01 g N2 -------------- 6.02*10²² molecules N2

( mass N2 ?? ) ----------- 25,000 molecules N2

mass N2 =  ( 25,000 * 28.01 ) /  ( 6.02*10²³ )

mass N2 = 700250 / 6.02*10²³

mass N2 = 1.163*10⁻¹⁸ g


Hope that helps!

7 0
2 years ago
Read 2 more answers
Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample
Dimas [21]

Answer:

<em>3.27·10²³ atoms of O</em>

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05\frac{g}{mol}.

We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.

19.3g <em>Na₂SO₄</em> · \frac{1 mol Na^2SO^4}{142.05g Na^2SO^4} · \frac{4 mol O}{1 mol Na^2SO^4} ·\frac{6.022x10^2^3}{1 mol O}

After doing the math for this dimensional analysis, you should get a quantity of approximately <em>3.27·10²³ atoms of O</em>.

3 0
2 years ago
The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T
kifflom [539]

Answer : The equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

Explanation :

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

First we have to calculate the standard free energy of reaction (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]

where,

\Delta G^o = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]

\Delta G^o=-90.72kJ/mol

Now we have to calculate the value of K_p

\Delta G^o=-RT\ln K_p

where,

\Delta G_^o =  standard Gibbs free energy  = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

K_p = equilibrium constant  = ?

Now put all the given values in this expression, we get:

-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p

K_p=7.98\times 10^{15}

Now we have to calculate the value of K_p.

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

The expression for equilibrium constant will be :

K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium SO_2 pressure be, x

Pressure of SO_2 = Pressure of H_2S = x

Now put all the given values in this expression, we get

7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}

x=3.93\times 10^{-5}torr

Thus, the equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

4 0
2 years ago
Chloe wants to lighten the color of her dark hair, which will require the use of both haircolor and hydrogen peroxide. hydrogen
bearhunter [10]

The answer is developer .

Chloe wants to lighten the color of her dark hair, which will require the use of both hair color and hydrogen peroxide.

Hydrogen peroxide (H₂O₂) is an example of a most commonly used developer in hair color.

Hydrogen peroxide lighten hair color, as it can dissolve darker pigment. According to the  natural hair color and the length of time a person will leave hydrogen peroxide on  hair,  hair can turn somewhere on the color spectrum between caramel, orange and yellow.





7 0
2 years ago
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