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3 years ago

WHEN parking uphill in a car with a manual transmission you should park with the transmission in

Chemistry

2 answers:

Stels [109]3 years ago

4 0

Answer:

In both cases, let the tire contact with the curb ease some of the pressure, apply the parking brake and then place the transmission into Park. If you have a manual transmission, place the transmission in first gear when parking facing downhill and reverse when parking facing uphill.

Explanation:

Alexandra [31]3 years ago

4 0

Answer:

first gear

Explanation: when parking a car uphill you use first but when parking a car downhill you use reverse.

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The acetic acid/acetate buffer system is a common buffer used in the laboratory. Write the equilibrium equation for the acetic a

hichkok12 [17]

Answer:

CH₃CO₂H + H₂O ⇄ CH₃CO₂⁻ + H₃O⁺

Explanation:

A buffer is defined as the mixture of a weak acid and its conjugate base or vice versa.

For the acetic acid buffer, CH₃CO₂H is the weak acid and its conjugate base is the ion without H⁺, that is CH₃CO₂⁻. The equilibrium equation in water knowing this is:

<em>In the equilibrium, the acid is dissociated in the conjugate base and the hydronium ion.</em>

8 0

3 years ago

Will give lots of points if answered correctly. Determine the kb for chloroform when 0.793 moles of solute in 0.758 kg changes t

Liono4ka [1.6K]

Answer: The value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

Explanation:

Given: Moles of solute = 0.793 mol

Mass of solvent = 0.758

$\Delta T_{b} = 3.80^{o}C$

As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.

$Molality = \frac{no. of moles}{mass of solvent (in kg)}\\= \frac{0.793 mol}{0.758 kg}\\= 1.05 m$

Now, the values of $K_b$ is calculated as follows.

$\Delta T_{b} = i\times K_{b} \times m$

where,

i = Van't Hoff factor = 1 (for chloroform)

m = molality

$K_{b}$ = molal boiling point elevation constant

Substitute the values into above formula as follows.

$\Delta T_{b} = i\times K_{b} \times m\\3.80^{o}C = 1 \times K_{b} \times 1.05 m\\K_{b} = 3.62^{o}C/m$

Thus, we can conclude that the value of $K_{b}$ for chloroform is $3.62^{o}C/m$ when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.

7 0

3 years ago

How do bacteria in the guts of cows help the cows to digest the grass they eat?

Andreyy89

A cow's<span> digestive system is quite different from that of humans. </span>Cows eat grass, hay and other plant material that contain hard-to-digest<span> cellulose. To cope with this </span>they<span> have a large stomach with four compartments, with the largest being the rumen.</span>

3 0

4 years ago

A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.9

Ronch [10]

Answer:

The free energy = -20.46 KJ

Explanation:

given Data:

Pb²⁺ = 0.750 M

Br⁻ = 0.232 M

R = 8.314 Jk⁻¹mol⁻¹

T = 298K

The Gibb's free energy is calculated using the formula;

ΔG = ΔG° + RTlnQ -------------------------1

Where;

ΔG° = standard Gibb's freeenergy

R = Gas constant

Q = reaction quotient

T = temperature

The chemical reaction is given as;

Pb²⁺(aq) + 2Br⁻(aq) ⇄PbBr₂(s)

The ΔG°f are given as:

ΔG°f (PbBr₂) = -260.75 kj.mol⁻¹

ΔG°f (Pb²⁺) = -24.4 kj.mol⁻¹

ΔG°f (2Br⁻) = -103.97 kj.mol⁻¹

Calculating the standard gibb's free energy using the formula;

ΔG° = ξnpΔG°(product) - ξnrΔG°(reactant)

Substituting, we have;

ΔG° =[1mol*ΔG°f (PbBr₂)] - [1 mol *ΔG°f (Pb²⁺) +2mol *ΔG°f (2Br⁻)]

ΔG° =(1 *-260.75 kj.mol⁻¹) - (1* -24.4 kj.mol⁻¹) +(2*-103.97 kj.mol⁻¹)

= -260.75 + 232.34

= -28.41 kj

Calculating the reaction quotient Q using the formula;

Q = 1/[Pb²⁺ *(Br⁻)²]

= 1/(0.750 * 0.232²)

= 24.77

Substituting all the calculated values into equation 1, we have

ΔG = ΔG° + RTlnQ

ΔG = -28.41 + (8.414*10⁻³ * 298 * In 24.77)

= -28.41 +7.95

= -20. 46 kJ

Therefore, the free energy of reaction = -20.46 kJ

8 0

3 years ago

There are several reagents that can be used to effect addition to a double bond including: acid and water, oxymercuration–demerc

ExtremeBDS [4]

The answer & explanation for this question is given in the attachment below.

7 0

3 years ago

Read 2 more answers