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Strike441 [17]
3 years ago
7

When solving a problem it is important to identify your given and needed units, but it is also important to understand the relat

ionship between those units so you will know how to set up your equation in order to solve the problem. Review the data sets below and use the steps of the problem-solving method to determine whether the given measurements would be appropriate for calculating mass, volume, or density.
a. 432 g of table salt occupies 20.0 cm^3 of space
b. 5.00 g 0T balsa wood, density of balsa wood : 0.16 g/cm^3
c. 32 cm^3 sample of gold density of gold 19.3 =g/cm^3
d. 150 g of iron, density of Iron = 79.0 g/cm^3
Chemistry
1 answer:
UNO [17]3 years ago
5 0

Answer:

See Explanation

Explanation:

Given

(a) to (d)

Required

Determine whether the given parameters can calculate the required parameter

To calculate either Density, Mass or Volume, we have

Density = \frac{Mass}{Volume}

Mass = Density * Volume

Volume = \frac{Mass}{Density}

(a) 432 g of table salt occupies 20.0 cm^3 of space

Here, we have:

Mass = 432g

Volume = 20.0cm^3

The above can be used to calculate Density as follows;

Density = \frac{Mass}{Volume}

Density = \frac{432g}{20.0cm^3}

Density = 21.6g/cm^3

(b) 5.00 g of balsa wood, density of balsa wood : 0.16 g/cm^3

Here, we have:

Mass = 5.00g

Density = 0.16g/cm^3

This can be used to solve for Volume as follows:

Volume = \frac{Mass}{Density}

Volume = \frac{5.00g}{0.16g/cm^3}

Volume = 31.25cm^3

(c) 32 cm^3 sample of gold density of 19.3 g/cm^3

Here, we have:

Volume = 32cm^3

Density = 19.3g/cm^3

This can be used to calculate Mass as follows:

Mass = Density * Volume

Mass = 32cm^3 * 19.3g/cm^3

Mass = 617.6g

(d) 150 g of iron, density of Iron = 79.0 g/cm^3

Here, we have

Mass = 150g

Density = 79.0g/cm^3

This can be used to calculate volume as follows:

Volume = \frac{Mass}{Density}

Volume = \frac{150g}{79.0g/cm^3}

Volume = 1.90 cm^3 <em>Approximated</em>

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En una estructura de concreto cuyo peso es de 8500 n se apoyo sobre un area de 25cm2,hallar la presion ejercida sobre su base
babymother [125]

Respuesta:

340 N/cm²

Explicación:

Paso 1: Información provista

Peso de la estructura (F): 8500 Newton

Area superficial (A): 25 cm²

Paso 2: Calcular la presión (P) ejercida por la estructura de concreto sobre su base

La presión es igual al cociente entre la fuerza ejercida y la superficie sobre la que se aplica.

P = F/A

P = 8500 N / 25 cm² = 340 N/cm²

5 0
3 years ago
What is the ratio of effusion rates for the lightest gas, h2, to the heaviest known gas, uf6?
andriy [413]

The ratio of effusion rates for the lightest gas H₂ to the heaviest known gas UF₆ is 13.21 to 1

<h3>What is effusion?</h3>

Effusion is a process by which a gas escapes from its container through a tiny hole into evacuated space.

Rate of effusion ∝ 1/√Ц, (where Ц is molar mass)

Rate H₂ = 1/√ЦH₂

Rate UF₆ = 1/√ЦUF₆

Therefore, Rate H₂/ Rate UF₆ = √ЦH₂/√ЦUF₆

ЦH₂= 2.016 g/mol

ЦUF₆= 352.04 g/mol

Rate H₂ / Rate UF₆ = √352.04/√2.016 = 18.76/1.42

Rate H₂ / Rate UF₆ = 13.21

Therefore, H₂ is lower mass than UF₆. Thus H₂ gas will effuse 13 times more faster than UF₆ because the most probable speed of H₂ molecule is higher; therefore, more molecules escapes per unit time.

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8 0
1 year ago
Which of the following statements is not an accurate description of a factor that contributes to the ordering of the spectrochem
MArishka [77]

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Explanation:

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5 0
3 years ago
Given K = 3.61 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.19 at 45°C for the reaction 2 A(g)
Firlakuza [10]

Answer:

K = 0.55

Kp = 0.55

mol fraction B = 0.27

Explanation:

We need to calculate the equilibrium constant for the reaction:

C(g) + D(g) ⇄ 2B(g)              K₁= ?                       (1)

and we are given the following equilibria with their respective Ks

A(g) + B(g) ⇄ C(g)                 K₂= 3.61                 (2)

2 A(g) + D(g)  ⇄ C(g)             K₃= 7.19                 (3)

all at 45 ºC.

What we need to do to solve this question is to manipulate equations (2) and (3)  algebraically  to get our desired equilibrium (1).

We are allowed to reverse  reactions, in that case we take the reciprocal of K as our new K' ; we can also  add two equilibria together, and the new equilibrium constant will be the product of their respective Ks .

Finally if we multiply by a number then we raise the old constant to that factor to get the new equilibrium constant.

With all this  in mind, lets try to solve our question.

Notice A is not in our goal equilibrium (3)  and we want D as a reactant . That  suggests we should reverse the first equilibria and multiply it by two since we have 2 moles of B  as product in our  equilibrium (1) . Finally we would add (2) and (3) to get  (1) which is our final  goal.

2C(g)             ⇄  2A(g) + 2B(g)  K₂´= ( 1/ 3.61 )²  

                                   ₊

2 A(g) + D(g)  ⇄     C(g)               K₃ = 7.19  

<u>                                                                                    </u>

C(g) + D(g)     ⇄    2B(g)       K₁ = ( 1/ 3.61 )²   x  7.19

                                             K₁ = 0.55

Kp is the same as K = 0.55 since the equilibrium constant expression only involves  gases.

To compute the last part lets setup the following mnemonic  ICE table to determine the quantities at equilibrium:

pressure (atm)        C             D           B

initial                     1.64          1.64         0

change                    -x             -x        +2x

equilibrium          1.64-x         1.64-       2x

Thus since

Kp =0.55 = pB²/ (pC x pD) = (2x)²/ (1.64 -x)²  where p= partial pressure

Taking square root to both sides of the equation we have

√0.55 = 2x/(1.64 - x)

solving for x  we obtain a value of 0.44 atm.

Thus at equilibrium we have:

(1.64 - 0.44) atm = 1.20 atm = pC = p D

2(0.44) = 0.88 = pB

mole fraction of B = partial pressure of B divided into the total gas pressure:

X(B) = 0.88 / ( 1.20 + 1.20 + 0.88 ) = 0.27

8 0
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Ymorist [56]

C.

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Explanation

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8 0
2 years ago
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