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2 years ago

15

Show proofs that the newspaper is perfectly reassembled

1 answer:

s2008m [1.1K]2 years ago

4 0

Answer : The pages of the newspaper contains many pictures which are perfectly assembled as we can see that, after the experiment was done, the words, the illustrations and the pictures in the original newspaper was actually formed using the cut clippings of newspaper.

The reassembly of most of the parts in the paper was displaying a readable text on top of the newspaper, which can be considered as main indicator of newspaper being perfectly assembled.

The physical appearance of the reassembled newspaper and the original newspaper was perfect and overall the almost similar to that of of old newspaper, it gave a clear indication as the newspaper was reassembled perfectly with no mistake in it.

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The middle block of the periodic table is called?

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Answer: The transition metals make up the middle block of he periodic table

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2 years ago

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2 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago

In a 28 g serving of cheese curls there are 247mg of sodium. How much sodium is in a 12.5 ounce bag

Marina86 [1]

Answer:

There should be about 3136.9mg of Na in a 12.5oz bag of cheese curls.when rounding if not rounded it would be 3126.04082143mg

Explanation:

12.5oz=354.4g

354.4/28=12.7

12.7*247=3136.9

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3 years ago

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