Ask question

Ask question

All categories

3 years ago

15

What is the specific heat of a metal with a mass of 14.0 g, heat of 3.45 kJ and a change in temperature of 3.2 ℃?

1 answer:

Zarrin [17]3 years ago

4 0

I think your question is not complete sir. supposely you can use Q=mc0.
(0.014)(4.2)(3.2)

You might be interested in

Brainliest!!!!!!Parking lots and other man made ground coverings may increase erosion around there edges

ollegr [7]

Answer:

true trust ong gotta be true fff

4 0

2 years ago

Tom looks very different from his parents he has brown eyes and his parents have blue eyes is he adopted??

SVEN [57.7K]

Answer:

No?

Explanation:

6 0

2 years ago

Read 2 more answers

ANSWER THESE 3 QUESTION AND I WILL GIVE OUT THE BRAINLIEST

Lana71 [14]

Answer:

A positive acceleration means and increase

A negative acceleration is a decrease

a zero acceleration is a constant movement so its not increasing or decreasing just constant

Explanation:

6 0

2 years ago

Read 2 more answers

(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

Notice what happens to the numbers of protons, neutrons, and electrons when you

babymother [125]

Answer:

Only the number of neutrons change.

3 0

3 years ago