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amm1812
3 years ago
14

A cube of plastic 1.2x10^-5 km on a side has a mass of 1.1 g. what is its density in g/cm^3

Chemistry
2 answers:
Vesna [10]3 years ago
7 0
Length of 1 side 1.2*10^-5km =1.2*10^-5*10^5 =1.2cm 

<span>volume of the cube (1.2)^3=1.728 cm^3 </span>

<span>density= mass/volume= 1.1/1.728=0.636 g/cm^3</span>
gizmo_the_mogwai [7]3 years ago
5 0

Answer:

0.6366  g/cm^3 is the density of the cubic plastic.

Explanation:

Mass of the cubic block= m = 1.1 g

Volume of the cubic block = V

Volume of the cube = a^3

a = side length = 1.2\times 10^{-5} km = 1.2 cm

1 km = 100,000 cm

Volume of cubic block = V=(12 cm)^3=1.728 cm^3

Density of the solid block,d = ?

Density=\frac{Mass}{Volume}

d=\frac{m}{V}=\frac{1.1 g}{1.728 cm^3}=0.6366 g/cm^3

0.6366 g/cm^3 is the density of the cubic plastic.

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point Deshawn ran the 400 m race(in a straight line) in 2 minutes. What is his distance and displacement ? Explain your answer b
san4es73 [151]

His distance and displacement are the same, which was 400 m

<h3>Further explanation</h3>

Given

Distance = 400 m

time = 2 min

Required

Distance and displacement

Solution

Distance is a scalar quantity that indicates the length of the trajectory that is traveled by an object within a certain interval. Distance has no direction, only has magnitude  

Can be simplified distance = totals traveled  

Displacement is a vector quantity that shows changes in the position of objects in a certain interval of time. Displacement has magnitude and direction  

Can be simplified displacement = distanced traveled from starting point to ending point  

From the definition above shows that the displacement and the distance that he traveled have the same value (magnitude), which is equal to 400 m

The value of the two will be different if he starts and finishes at the same point, then the displacement value is zero while the distance he has traveled is still 0

4 0
3 years ago
The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed
Oksanka [162]

Answer:

(a) The rate of disappearance of O_{2} is: 4.65*10^{-5} M/s

(b) The value of rate constant is: 0.83036 M^{-2}s^{-1}

(c) The units of rate constant is:  M^{-2}s^{-1}

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

2NO(g)+O_{2}->2NO_{2}

rate = -\frac{1}{2} \frac{d}{dt}[NO] = -\frac{d}{dt}[O_{2}] = \frac{1}{2}\frac{d}{dt}[NO_{2}] -----(1)

According to the question, the reaction is second order in NO and first order in  O_{2}.

Then we can say that, rate = k[NO]^{2}[O_{2}] -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

-\frac{d}{dt}[NO] = 9.3*10^{-5} M/s.

(a) From (1), we can get the rate of disappearance of O_{2}.

    Rate of disappearance of  O_{2} = -\frac{d}{dt}[O_{2}] = (0.5)*(9.3*10^{-5}) M/s = 4.65*10^{-5} M/s.

(b) The rate of the reaction can be obtained from (1).

    rate = -\frac{1}{2} \frac{d}{dt}[NO] = (0.5)*(9.3*10^{-5})

    rate = 4.65*10^{-5} M/s

   The value of rate constant can be obtained by using (2).

    rate constant = k = \frac{rate}{[NO]^{2}[O_{2}]}

    k = \frac{4.65*10^{-5}}{(0.040)^{2}(0.035)} = 0.83036 M^{-2}s^{-1}

(c) The units of the rate constant can be obtained from (2).

    k = \frac{rate}{[NO]^{2}[O_{2}]}

    Substituting the units of rate as M/s and concentrations as M, we get:

\frac{Ms^{-1} }{M^{3}} = M^{-2}s^{-1}

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

     rate\alpha [NO]^{2}

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of (1.8)^{2} = 3.24

     

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3 years ago
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Yanka [14]
Every substance is either an element or a compound.
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