All categories

3 years ago

A car gets 28.0 miles/gallon of gas on the highway. What is this in km/L?

Chemistry

1 answer:

lys-0071 [83]3 years ago

4 0

Answer:

11.905 km/L

Explanation:

Given that,

A car gets 28.0 miles/gallon of gas on the highway.

We need to find the volume into km/L

We know that,

1 mile = 1.60934 km

Also,

1 gallon = 3.785 litre

So,

$28\ \text{miles per gallon}=28\times \dfrac{1.60934\ km}{3.785\ L}\\\\=11.905\ \text{km/L}$

Hence, the car get 11.905 km/L of the gas.

You might be interested in

A system fitted with a piston expands when it absorbs 53.1 ) of heat from the surroundings. The piston is working against a pres

Hoochie [10]

Answer:

d. 60.8 L

Explanation:

Step 1: Given data

Heat absorbed (Q): 53.1 J

External pressure (P): 0.677 atm

Final volume (V2): 63.2 L

Change in the internal energy (ΔU): -108.3 J

Step 2: Calculate the work (W) done by the system

We will use the following expression.

ΔU = Q + W

W = ΔU - Q

W = -108.3 J - 53.1 J = -161.4 J

Step 3: Convert W to atm.L

We will use the conversion factor 1 atm.L = 101.325 J.

-161.4 J × 1 atm.L/101.325 J = -1.593 atm.L

Step 4: Calculate the initial volume

First, we will use the following expression.

W = - P × ΔV

ΔV = - W / P

ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L

The initial volume is:

V2 = V1 + ΔV

V1 = V2 - ΔV

V1 = 63.2 L - 2.35 L = 60.8 L

7 0

3 years ago

PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Masteriza [31]

Answer:

$\large \boxed{\text{-827.4 kJ}}$

Explanation:

We have three equations:

1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ

3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.

7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

6. 2H₂S(g) + O₂(g) ⟶ 2S(s) + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

$\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}$

8 0

3 years ago

TUESDAY:

Rzqust [24]

Protons are positive
Neutrons are neutral
Electrons are negatively charged

5 0

4 years ago

Read 2 more answers

What types of objects are using electricity in your home today?

EleoNora [17]

Lamp, tv, refrigerator, Lights, phone, microwave, surround sound, cameras, motion sensors, chandelier, pool, hot tub, heated floors, the smart home stuff, aquarium stuff, and whatever the staff uses.

3 0

3 years ago

A pulse with an amplitude of 3+ would be considered:

alexdok [17]

A pulse with an amplitude of 3+ would be considered as increased. Peripheral Pulse Assessment Grading System is measured in 0 - 3 Scale. 0 = absent 1+ = Weak/thready pulse 2+ Normal Pulse 3+ = Full, firm pulse. from the above scale we can find that the 3+ reading shows that the pulse is increased.

4 0

4 years ago