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mario62 [17]
3 years ago
12

Write the oxidation number for the following elements: for brainliest answer

Chemistry
1 answer:
ollegr [7]3 years ago
6 0

Explanation:

Here's an oxidation chart to help

..................

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what is the balanced equation for when an aqueous solution of hydrochloric acid reacts with sodium metal, aqueous sodium chlorin
FinnZ [79.3K]

Answer:

2Na(s) + 2HCl(aq) —> 2NaCl(aq) + H2(g)

Explanation:

Na(s) + HCl(aq) —> NaCl(aq) + H2(g)

Writing an ionic equation will actually help us to understand the equation and also to balance it. This is illustrated below:

Na + H+Cl-

Na is higher than H in the activity series and as such, it will displaces H from the solution and form NaCl with H2 liberated as shown below

Na + H+Cl- —> Na+Cl- + H2

Now, put 2 in front of Na, H+Cl- and Na+Cl- to balance the equation as shown below:

2Na + 2H+Cl- —> 2Na+Cl- + H2

Now we can write the elemental equation as follow:

2Na(s) + 2HCl(aq) —> 2NaCl(aq) + H2(g)

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2 years ago
Which of these phrases would go in the overlap? Select two options.
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I think it’s found in the nucleus and has a mass of one amu
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a normal hemoglobin concentration in the blood is 15g/100ml of blood how many kilogram of hemoglobin are in a person who has 5.5
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Using stoichiometry:

5.5 L of blood x (1000 mL/1L) x (15 g/100 mL) x (1 kg/1000 g) = 0.825 kg
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When 0.100 mol of carbon is burned in a closed vessel with8.00
antoniya [11.8K]

Answer : The mass of carbon monoxide form can be 2.8 grams.

Solution : Given,

Moles of C = 0.100 mole

Mass of O_2 = 8.00 g

Molar mass of O_2 = 32 g/mole

Molar mass of CO = 28 g/mole

First we have to calculate the moles of O_2.

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{8g}{32g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2C+O_2\rightarrow 2CO

From the balanced reaction we conclude that

As, 2 mole of C react with 1 mole of O_2

So, 0.1 moles of C react with \frac{0.1}{2}=0.05 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and C is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO

From the reaction, we conclude that

As, 2 mole of C react to give 2 mole of CO

So, 0.1 moles of C react to give 0.1 moles of CO

Now we have to calculate the mass of CO

\text{ Mass of }CO=\text{ Moles of }CO\times \text{ Molar mass of }CO

\text{ Mass of }CO=(0.1moles)\times (28g/mole)=2.8g

Therefore, the mass of carbon monoxide form can be 2.8 grams.

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Corrosion may be regarded as the destruction of metal by:
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