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Tcecarenko [31]

3 years ago

9

12. Contrast the movement of charges in a

1 answer:

nikdorinn [45]3 years ago

4 0

DEEZ NUTS ha gattti -sorry

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Iron forms rust, expressed as Fe2O3.

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Answer:

it's C. 159.70 g

Explanation:

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What are glands.......................

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An organ in the human or animal body that secretes particular chemical substances for use in the body or for discharge into the surroundings.

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A. What type of elements are found in each group?

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Groups are numbered 1–18 from left to right. The elements in group 1 are known as the alkali metals; those in group 2 are the alkaline earth metals; those in 15 are the pnictogens; those in 16 are the chalcogens; those in 17 are the halogens; and those in 18 are the noble gases.

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Heart my answer if it's helpful

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3 years ago

Some distilled water is added to an empty beaker. a gram of copper (ii) nitrate is added to the beaker and while the water is be

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3 years ago

A lab scientist cools a liquid sample of water (2.6 kg) at 0.00°C to -192°C. The water turns to ice as this temperature change o

soldier1979 [14.2K]

<u>Answer:</u> The heat released for the given process is -1892 kJ

<u>Explanation:</u>

The processes involved in the given problem are:

$1.)H_2O(l)(0^oC,273K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(s)(-192^oC,81K)$

Pressure is taken as constant.

To calculate the amount of heat released at same temperature, we use the equation:

$q=m\times L_{f,v}$ ......(1)

where,

q = amount of heat released = ?

m = mass of water/ice

$L_{f,v}$ = latent heat of fusion or vaporization

To calculate the amount of heat released at different temperature, we use the equation:

$q=m\times C_{p,m}\times (T_{2}-T_{1})$ .......(1)

where,

q = amount of heat released = ?

$C_{p,m}$ = specific heat capacity of medium

m = mass of water/ice

$T_2$ = final temperature

$T_1$ = initial temperature

Calculating the heat absorbed for each process:

<u>For process 1:</u>

Converting the latent heat of fusion in J/kg, we use the conversion factor:

1 kJ = 1000 J

So, $(\frac{-334kJ}{1kg})\times (\frac{1000J}{1kJ})=-334\times 10^3J/kg$

We are given:

$m=2.6kg\\L_f=-334\times 10^3J/kg$

Putting values in equation 1, we get:

$q_1=2.6kg\times (-334\times 10^3J/kg)=-868400J$

<u>For process 2:</u>

We are given:

$m=2.6kg\\C_{p,s}=2050J/kg.K\\T_1=273K\T_2=81K$

Putting values in equation 2, we get:

$q_2=2.6kg\times 2050J/kg.K\times (81-(273))^oC\\\\q_2=1023360J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[-868400+(-1023360)]J=-1891760J=-1891.76kJ\approx -1892kJ$

Hence, the heat released for the given process is -1892 kJ

3 0

3 years ago