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3 years ago

A/1 * B/C =D

Chemistry

1 answer:

Korvikt [17]3 years ago

7 0

Exchange rate is 1.5 guilders for every 2.4 florins

The conversion factor will be

1.5 guilders / 2.4 florins = 2.4 florins / 1.5 guilders = 1

so if we wish to convert 4 guilders to florins we will use following conversion factor

2.4 florins / 1.5 guilders = B / C

D = 4guilders X 2.4 florins / 1.5 guilders = 6.4 florins

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PLEASE HELP DUE SOON

Alona [7]

Answer:

Because energy is conserved, the kinetic energy of a block at the bottom of a frictionless

ramp is equal to the gravitational potential energy of the block at the top of the ramp. This

value is proportional to the square of the block’s velocity at the bottom of the ramp.

Therefore, the block’s final velocity depends on the height of the ramp but not the steepness

of the ramp

Explanation:

7 0

2 years ago

The rate constant for a certain reaction is k = 4.70×10−3 s−1 . If the initial reactant concentration was 0.700 M, what will the

Lynna [10]

Answer:

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

Explanation:

The unit of k is s⁻¹.

The order of the reaction = first order.

First order reaction: A first order reaction is a reaction in which the rate of reaction depends only the value of the concentration of the reactant.

$-\frac{d[A]}{dt} =kt$

[A] = the concentration of the reactant at time t

k= rate constant

t= time

Here k= 4.70×10⁻³ s⁻¹

t= 4.00

[A₀] = initial concentration of reactant = 0.700 M

$-\frac{d[A]}{dt} =kt$

$\Rightarrow -\frac{d[A]}{[A]}=kdt$

Integrating both sides

$\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt$

⇒ -ln[A] = kt +c

When t=0 , [A] =[A₀]

-ln[A₀] = k.0 + c

⇒c= -ln[A₀]

Therefore

-ln[A] = kt - ln[A₀]

Putting the value of k, [A₀] and t

- ln[A] =4.70×10⁻³×4 -ln (0.70)

⇒-ln[A]= 0.375

⇒[A] = 0.686

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

5 0

3 years ago

which of the following results when iron and oxygen react? 1.fe2o3 corrodes 2.hydrated iron forms 3.iron oxide forms 4.a shiny m

weeeeeb [17]

It should be 4, a shiny metal forms

3 0

3 years ago

) Given the following balanced equation, determine the rate of reaction with respect to [O2]. If the rate of formation of O2is 7

Travka [436]

Answer:

Rate of the reaction is 0.2593 M/s

-0.5186 M/s is the rate of the loss of ozone.

Explanation:

The rate of the reaction is defined as change in any one of the concentration of reactant or product per unit time.

$2O_3\rightleftharpoons 3O_2$

Rate of formation of oxygen : $7.78\times 10^{-1} M/s$

Rate of the reaction(R) = $\frac{-1}{2}\frac{d[O_3]}{dt}=\frac{1}{3}\frac{d[O_2]}{dt}$

$R=\frac{1}{3}\frac{d[O_2]}{dt}$

Rate of formation of oxygen=3 × (R)

$7.78\times 10^{-1} M/s=3\times R$

Rate of the reaction(R): $0.2593 M/s$

Rate of the reaction is 0.2593 M/s

Rate of disappearance of the ozone:

$R=-\frac{1}{2}\frac{d[O_3]}{dt}$

$\frac{d[O_3]}{dt}=-2\times R=-2\times 0.2593\times M/s=-0.5186M/s$

-0.5186 M/s is the rate of the loss of ozone.

6 0

3 years ago

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago