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Delvig [45]
3 years ago
6

A car traveling with constant speed travels 150 km in 7200 s. What is the speed of the car?

Chemistry
1 answer:
Katena32 [7]3 years ago
5 0
Mass divided by the time = speed

mass = 150 km
time = 7200 s

Divide

7200/150 = s per km
48 seconds per km

Answer: 48 m/s

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Something with a density greater than 1.00 g ml will sink in waterThe density of water is 1 gm/cm^3=1000 kg/m^3, therefore, any element or object with density greater than 1 will sink in water.
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Nickel ( 8910 kg/m^3 = 8.91 gm/cm^3)
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Propose a 3-step reaction for the conversion of cyclopentene to propanedioic acid
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Alkynes are reduced to trans alkenes by a process called dissolving metal reduction. The reaction uses sodium or lithium metal a
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Explanation:

The movement of the electrons is illustrated in the picture attached to this answer. It is a four-step reaction mechanism.

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Second STEP: This radical anion then removes a proton/hydrogen from ammonia in a bid to neutralize itself (hence the hydrogen becomes bonded to the anion).

Third STEP: The sodium (from NaNH₂ formed) transfers an electron again to produce a vinyl carbanion.

Fourth STEP: The carbanion then removes a proton/hydrogen from ammonia (like in the second step) to form a neutral trans-alkene.

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6 0
3 years ago
Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under eac
valentinak56 [21]

Answer : The value of \Delta G_{rxn} is, 8.867kJ/mole

Explanation :

The formula used for \Delta G_{rxn} is:

\Delta G_{rxn}=\Delta G^o+RT\ln Q   ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction

\Delta G_^o =  standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

T = temperature = 25^oC=273+25=298K

Q = reaction quotient

First we have to calculate the \Delta G_^o.

Formula used :

\Delta G^o=-RT\times \ln K_p

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

\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})

\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

CO(g)+2H_2(g)\rightarrow CH_3OH(g)

The expression for reaction quotient will be :

Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}

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

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

Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}

Q=8.1\times 10^{5}

Now we have to calculate the value of \Delta G_{rxn} by using relation (1).

\Delta G_{rxn}=\Delta G^o+RT\ln Q

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

\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})

\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole

Therefore, the value of \Delta G_{rxn} is, 8.867kJ/mole

3 0
3 years ago
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