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3 years ago

The unit conversion between rankine and kelvin is (linear, exponential, quadratic, none of the above)?

Chemistry

1 answer:

diamong [38]3 years ago

4 0

Answer:

Unit conversion between Rankine and Kelvin is linear.

Unit conversion between degree Celsius and degree Fahrenheit is linear.

Explanation:

Relation between rankine and Kelvin is

$R\ =\ \dfrac{9}{5}\ K$

So, the plot between Rankine and Kelvin is a straight line with zero intercept and has a slope having value $\dfrac{9}{5}$ .

Relation between degrees Celsius and degrees Fahrenheit is given by

$^{\circ}C\ =\ \dfrac{5}{9}\ (^{\circ}F\ -\ 32)$

So, the plot between degrees Celsius and degrees Fahrenheit is a straight line with slope $\dfrac{5}{9}$ and negative intercept of $\dfrac{160}{9}$ .

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An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its

german

Hello!

First you need to calculate q
delta U is change in internal energy 

delta U = q + w 
q is heat and w work done 
here work was done by the system means energy leaving the system so w is negative 

delta U = q + w 

q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ 

q = m x c x delta T 

7211 J = 80.0 g x c x (225-25) °C 

c = 0.451 J /g °C

Hope this Helps! Have A Wonderful Day! :)

5 0

2 years ago

Q1.

dlinn [17]

Answer:

there are four peaks in the 13c NMR (B)

3 0

2 years ago

Question 34 (1 point)

Grace [21]

Answer:

8.33 atm

Explanation:

Xe is 5 out of (4+5) or 5 / 9 ths of the gas present

5/9 * 15 atm = 8.33 atm

4 0

1 year ago

Limiting Reactants—————-

denis-greek [22]

Answer:

21.8 grams.

Explanation:

Molar mass data from a modern periodic table:

Mg: 24.301;
O: 15.999.

How many moles of MgO will be produced if Mg is the limiting reactant?

Number of moles of Mg:

$\displaystyle n = \frac{m}{M} = \frac{16.3}{24.301} = 0.670644\;\text{mol}$ .

The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.

How many moles of MgO will be produced if O₂ is the limiting reactant?

Number of moles of O₂:

$\displaystyle n = \frac{m}{M} = \frac{4.33}{15.999} = 0.270642\;\text{mol}$ .

The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO. $2\times 0.270642 = 0.541284\;\text{mol}$ of MgO will be produced if O₂ is in excess.