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AlladinOne [14]
2 years ago
5

What is the value of c? 56.70

Chemistry
1 answer:
Gennadij [26K]2 years ago
6 0

Due to the lack of information provided, the best I can give you is this:

In Chemistry, the letter c most commonly refers to specific heat, which is present in the ratio J/g°C.

Hope it helps :)

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lutik1710 [3]

Answer:

hey ifk

Explanation:

5 0
2 years ago
The reaction 2a + b --> c + d has an activation energy of 80.0 kj/mol. at 320°c, the rate constant k = 1.80 x 10-2 l mol-1 s-
Naddik [55]
To determine the k for the second condition, we use the Arrhenius equation which relates the rates of reaction at different temperatures. We do as follows:

ln k1/k2 = E / R (1/T2 - 1/T1) where E is the activation energy and R universal gas constant.

ln 1.80x10^-2 / k2 = 80000 / 8.314 ( 1/723.15 - 1/593.15)

k2 = 0.3325 L / mol-s
5 0
3 years ago
(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe
Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration C_o = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

time (t) = \dfrac{V}{Q}

time (t) = \dfrac{200}{0.3}

time (t) = 666.66 \ sec

time (t) = \dfrac{666.66 }{3600} hrs

time (t) = 0.185 hrs

\text{Using First Order Reaction:}

\dfrac{dc}{dt}=kc

where;

t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)

0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)

0.942 =  \Big ( \dfrac{200}{C_e}- 1 \Big)

1+ 0.942 =  \Big ( \dfrac{200}{C_e} \Big)

\dfrac{200}{C_e} = 1.942

C_e = \dfrac{200}{1.942}

\mathbf{C_e = 102.98 \ mg/l}

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]

\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)

\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)

\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)

V_{PFR} =0.196(0.663)*0.3*3600

V_{PFR} = 140.34 \ m^3

The volume of the PFR is ≅ 140 m³

3 0
3 years ago
How many grams of a 26.9% sugar solution contain 49.0 g of sugar?
Ronch [10]

Answer:

182.156g

Explanation:

grams = 49/.269 = 182.156g needed

6 0
2 years ago
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
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