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4 years ago

In an oxidation half-reaction, which amount is shown?

Chemistry

2 answers:

Advocard [28]4 years ago

5 0

Answer: option C

Explanation:

Oxidation means you loss of electons (LEO) and Reduction is gain of electrons.

Example: sodium gets oxidized by the loss of one electron as shown below

Na------> Na+ + 1e-

Here Na lost one electron hence it is oxidation

Change in oxidation no: 0 to +1

Tatiana [17]4 years ago

4 0

Hi,
The answer should be C.

Hope this helps, if you’d like further explanation please let me know.

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Meiosis happens in the major cells of multicellular organisms. true or false

olga55 [171]

Answer:

True.

Explanation:

Btw ya spelled mitosis wrong...

8 0

4 years ago

a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat

Nostrana [21]

Answer:- $\frac{382.69J}{0C}$ .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, $\Delta T$ for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

$q=mc\Delta T$

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

$q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})$

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

$\Delta T$ for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

$q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)$

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, $q=C.\Delta T$

$C=\frac{q}{\Delta T}$

$C=\frac{1044.746J}{2.73^0C}$

$C=\frac{382.69J}{0C}$

So, the heat capacity of calorimeter is $\frac{382.69J}{0C}$ .

4 0

3 years ago

Limestone is made of calcium carbonate, CaCO3. Would you use an acid or a base to test a rock sample to determine whether it con

mote1985 [20]

Answer:We will use an acid to test whether sample contains calcium carbonate or not.

Explanation:

Calcium carbonate reacts with acids to give off carbon-dioxide gas ,water and calcium chloride. it is a salt of strong base calcium hydroxide and weak acid carbonic acid.

$CaCO_3+2HCl\rightarrow CO_2+H_2O+CaCl_2$