3. How fast will her 1500 kg car accelerate if it produces 5000 N of force?

1.5mol C3H8 from C3H8+5O2-->3CO2+4H2O .how many grams of carbon dioxide are produced

koban [17]

Answer:

$\large \boxed{\text{200 g CO}_{{2}}}$

Explanation:

We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.

Mᵣ: 44.01

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

n/mol: 1.5

1. Calculate the moles of CO₂

The molar ratio is 3 mol CO₂:1 mol C₃H₈

$\rm \text{Moles of CO}_{2} = \text{1.5 mol C$_{3}$H}_{8} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C$_{3}$H}_{8}} =\text{4.5 mol CO}_{2}$

2. Calculate the mass of CO₂.

$\text{Mass of CO}_{2} = \text{4.5 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO$_{2}$}} = \textbf{200 g CO}_{\mathbf{2}}\\\text{The reaction will form $\large \boxed{\textbf{200 g CO}_{\mathbf{2}}}$}$

3 0

2 years ago

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

Answer: The time taken by the reaction is 84.5 seconds

Explanation:

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

6 0

3 years ago

An object was measured by a worker as

Alexeev081 [22]

Answer:.633

Explanation:

I have know idea but it was right

3 0

2 years ago

The smell of fresh cut pine is due in part to the cyclic alkene called pinene. Given the following data of pinene: Vapor pressur

Sveta_85 [38]

Answer: Heat of vaporization is 41094 Joules

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = initial pressure at 429 K = 760 torr

$P_2$ = final pressure at 415 K = 515 torr

= enthalpy of vaporisation = ?

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 429 K

$T_2$ = final temperature = 515 K

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

$\log (\frac{515}{760}=\frac{\Delta H}{2.303\times 8.314J/mole.K}[\frac{1}{429K}-\frac{1}{415K}]$

$\Delta H=41094J$

Thus the heat of vaporization is 41094 Joules

4 0

2 years ago

1)

Delicious77 [7]

Answer:

C) How much energy was added to the substance to increase molecule motion?

Explanation:

The most relevant question to ask regarding this change must take into account the physical knowledge about matter.

When matter changes from liquid state to gaseous state, a physical change called evaporation, the particles (molecules or atoms) of the pure substance will separate from each other, take up more space and move faster.

Condensation is the opposite to evaporation, thus the option A) is not the most relevant question.

The charge of the particles does not change; so the option B) is not relevant at all.

The particles should gain energy from the surroundings to increase their motion (kinetic energy) when they pass from liquid state, where they move slower, to gas state, where they move faster. Hence, the option C), How much energy was added to the substance to increase molecule motion? , is totally relevant.

Since this is an increase in the kinetic energy of the molecules, the option D) is not relevant.

6 0

3 years ago

3. How fast will her 1500 kg car accelerate if it produces 5000 N of force?​

3. How fast will her 1500 kg car accelerate if it produces 5000 N of force?