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3 years ago

13

Using the formula for velocity shown below, what is the average velocity of a bicyclist who biked 20 miles north of where he beg

an for a duration of 4 hours?

1 answer:

Marat540 [252]3 years ago

3 0

8.1km/hr due North

Explanation:

Given parameters:

Displacement = 20 miles North

Time = 4hours

Unknown;

Velocity = ?

Solution:

Velocity is the speed in a particular direction. It is simply displacement divided by time. A change in position with time.

Velocity = $\frac{displacement}{time}$

The unit is usually expressed in km/hr or m/s

We need to convert the given displacement to that in km:

1 mile = 1.609km

20miles = 1.609 x 20, 32.18km

Velocity = $\frac{32.18}{4}$ = 8.1km/hr due North

Learn more:

Velocity brainly.com/question/10962624

#learnwithBrainly

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The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

Balance the following reaction in KOH (under basic conditions). What are the coefficients in for C3H8O2 and KMnO4 in the balance

GrogVix [38]

Answer:

Coefficient of $C_3H_8O_2=3$

Coefficient of $KMnO_4$ =8

Explanation:

We are given that a reaction in which $C_3H_8O_2$ reacts with $KMnO_4$

We have to find the coefficient of each reactants in balanced reaction

$3C_3H_8O_2(aq)+8KMnO_4(aq)\rightarrow 3C_3H_2O_4K_2(aq)+8MnO_2(aq)+2KOH+8H_2O$

Coefficient is defined the constant value multiplied with a reactant in a reaction.

Coefficient of $C_3H_8O_2$ =3

Coefficient of $KMnO_4=8$

Coefficient of $C_3H_2O_4K_2=3$

Coefficient of $MnO_2=8$

Coefficient of $H_2O=8$

Coefficient of KOH=2

Hence, Coefficient of $C_3H_8O_2=3$ and coefficient of $KMnO_4=8$

7 0

3 years ago

anyanavicka [17]

I believe this a PV = nRT question whereas

you re write the formula and solve for volume

V = nRT/ P

then you input the values

P= pressure constant

V= x

n = moles = 0.2540

R = gas constant should be 8.314J mol

T = C degrees + 273.15 = K

solve for voume

make sure all units match

and use sig figs!!!!

7 0

3 years ago

At a certain temperature the vapor pressure of pure methanol is measured to be . Suppose a solution is prepared by mixing of met

CaHeK987 [17]

Answer:

Partial pressure is 0.13 atm

Explanation:

CHECK THE COMPLETE QUESTION BELOW :

At a certain temperature, the vapor pressure of pure methanol is measured to be 0.43atm. Suppose a solution is prepared by mixing 88.2 g of methanol and 116.g of water. Calculate the partial pressure of methanol vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal.

Using Raoult´s law for ideal soultions we have

P(A) = X(A) *Pº(A)

where P(A) is the partial vapor pressure pressure of methanol,

X(A) is the mole fraction of solute (methanol) in solution,

Pº(A) is the vapor pressure of pure solute

Raoult's law states that the vapor pressure of a solution is dependent on the mole fraction of a solute added to the solution.

Raoult's law can be expressed below

Psolution = ΧsolventP0solvent.

Expressing it interns of the constituents given in the question we have

P(CH₃OH) = X(CH₃OH) x Pº(CH₃OH)

To calculate the mole fraction of CH₃OH, we make use of the formula below :

X(A) = mol (A) / ntotal

Ntotal = (sum of number of moles of A )+( moles solvent)

mol (CH₃3OH) can be calculated as :: 88.2 g/ 32 g/mol = 2.76 mol of CH₃OH

mol (H₂O) can be calculated as ::116 g/ 18 g/mol = 6.44 mol

total n = (6.44 + 2.76) mol = 9.20 mol

To calculate the partial pressure the we say;

P(CH₃OH) = (2.76 mol CH + 9.20 mol) x( 0.43 atm) = 0.13 atm

Hence, the partial pressure rounded to two significant figures is 0.13 atm

8 0

3 years ago

2c2h6 + 7o2 = 4co2 + 6h2o <br> what mass of co2 will be produced from the reaction of 37.5 g c2h6

ivanzaharov [21]

Answer:

use n=m/M (moles=mass/molar mass) to find out how many moles of C2H6 there are in 60g

Explanation:

8 0

3 years ago