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xz_007 [3.2K]
2 years ago
6

A chemist obtains a sample of

Chemistry
1 answer:
djverab [1.8K]2 years ago
7 0

The number of formula units of Na₂SO₄ present in the sample is 3.01×10²³ units

<h3>Avogadro's hypothesis </h3>

From Avogadro's hypothesis,

1 mole of Na₂SO₄ = 6.02×10²³ units

But

1 mole of Na₂SO₄ = (23×2) + 32 + (16×4) = 142 g

Thus,

142 g of Na₂SO₄ = 6.02×10²³ units

<h3>How to determine the units in 71 g of sample </h3>

142 g of Na₂SO₄ = 6.02×10²³ units

Therefore,

71 g of Na₂SO₄ = (71 × 6.02×10²³) / 142

71 g of Na₂SO₄ = 3.01×10²³ units

Thus, 3.01×10²³ units of Na₂SO₄ is present in the sample

Learn more about Avogadro's number:

brainly.com/question/26141731

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Element      atomic number       position

Ba              56                             group 2, period 6

Ca              12                             group 2, period 3

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Now, you need to know the properties of the different type of elements and the tendencies on the periodic table.


The metallic elements are, those placed on the left side of the periodic table, are the ones that release an electron more easily, so they will requiere less energy to give it up when forming chemical bonds.


The higher the metallic character the less the energy need to give up an electron.


The metallic character grows as the group number decreases (goes to the left) period increases (goes downward), so among the  elements considered, Barium will require the least amount of energy to give un an electron when forming chemical bonds.
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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
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Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

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