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3 years ago

15

Can anyone help me with this?

1 answer:

satela [25.4K]3 years ago

4 0

Answer:liquids and gases

Explanation:

friction

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Are ions always smaller than their neutral atoms

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The negative ion is larger than the original atom. but the same number of positive protons, the size of the ionic radius will increase. When an atom has its electrons attracted to another atom it becomes a positive ion. The positive ion is smaller than the original atom.

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The intermolecular forces of attraction in hydrogen gas are stronger than those of helium

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Balance the equation.<br>P + 02 --> P4O10

charle [14.2K]

Answer:

4P + 5O₂ —> P₄O₁₀

Explanation:

From the question given above, we obtained:

P + O₂ —> P₄O₁₀

The above equation can be balance as illustrated below:

P + O₂ —> P₄O₁₀

There are 4 atoms of P on the right side and 1 atom on the left side. It can be balance by 4 in front of P as shown below:

4P + O₂ —> P₄O₁₀

There are 10 atoms of O on the right side and 2 atoms on the left side. It can be balance by putting 5 in front of O₂ as shown below:

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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

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3 years ago