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2 Na + Cl2 --> 2 NaCl<br> What is the mole ratio of sodium to sodium chloride?

Irina18 [472]

Answer: The mole ratio of sodium to sodium chloride 2:2.

Explanation:

As the given reaction equation is as follows.

$2Na + Cl_{2} \rightarrow 2NaCl$

Here, 2 moles of sodium reacts with 1 mole of $Cl_{2}$ and leads to the formation of 2 moles of NaCl.

This means that 2 moles of sodium gives 2 moles of NaCl on reaction with chlorine.

Hence, the ratio of moles of sodium to sodium chloride is 2:2.

Thus, we can conclude that the mole ratio of sodium to sodium chloride 2:2.

6 0

3 years ago

Introduced sewage in water systems _______. a. decreases levels of phosphorus and nitrogen b. increases oxygen levels c. increas

enyata [817]

Option C: Introducing sewage in water systems increases the concentration of solids.

Sewage is one of the most concerning sources of water pollution. Sewage, in simple terms, is the wastewater that contains compostable materials from the kitchen, industries, bathrooms, etc.

When untreated sewage is introduced into the water bodies, the microorganisms in it take up the oxygen from water for their aerobic respiration. As a result, the resident organisms of the water body have to struggle to survive due to the lack of oxygen in the water.

Sewage also contributes to the eutrophication of water bodies. Eutrophication is the process by which the nutritional elements, like phosphates and nitrates, in a water body, are enhanced causing the microorganisms to feast over them.

Sewage has some solid particles that can't be dissolved or decomposed. This then leads to the concentration of solids at the base of the water body. This further leads to the growth of various organisms using dissolved oxygen and leads to the deterioration of water quality.

To know more about sewage water, refer to the following link:

brainly.com/question/10022625

#SPJ4

8 0

2 years ago

Gasoline is a mixture of compounds. A gasoline sample was found to be 2.851% (by mass) octane (C₈H₁₈). 482.6 g of gasoline is co

Pani-rosa [81]

Answer:

See below.

Explanation:

The mass of octane in the sample of gasoline is 0.02851 * 482.6 = 13.759 g of octane.

The balanced equation is:

2C8H18(l) + 25O2(g) ----> 16CO2(g) + 18H2O(g)

From the equation, using atomic masses:

228.29 g of octane forms 704 g of CO2 and 324.3 g of H2O

So the mass of CO2 formed from the combustion of 13.759 g of octane = (704 * 13.759) / 228.29

= 42.43 g of CO2.

Amount of water = 324.3 * 13.759) / 228.29

= 19.55 g of H2O.

5 0

3 years ago

What is the approximate mass of a 100-N person on earth

snow_lady [41]

It would be 10kg, because 1kg = 10N

8 0

4 years ago

g An oxidized silicon (111) wafer has an initial field oxide thickness of d0. Wet oxidation at 950 °C is then used to grow a thi

77julia77 [94]

Answer:

Explanation:

From the information given:

oxidation of oxidized solution takes place at 950° C

For wet oxidation:

The linear and parabolic coefficient can be computed as:

$\dfrac{B}{B/A} = D_o \ exp \Big [\dfrac{-\varepsilon a}{k_BT} \Big]$

Using $D_o$ and $E_a$ values obtained from the graph:

Thus;

$\dfrac{B}{A} = 1.63 \times 10^8 exp \Big [ \dfrac{-2.05}{8.617 \times 10^{_-5}\times 1173}\Big] \\ \\ = 0.2535 \ \ \mu m/hr$

$B= 386 \ exp \Big [-\dfrac{0.78}{8.617 \times 10^{-3} \times 1173} \Big] \\ \\ = 0.1719 \ \mu m^2/hr$

So, the initial time required to grow oxidation is expressed as:

$t_{ox} = \dfrac{x}{B/A}+ \dfrac{x^2}{B} - t_o (initial)$

$where; \\ \\ t_{ox} = 2 \ hrs;\\ \\ x = 0.5 \\ \\ B/A = 0.2535 \\ \\ B = 0.1719$

∴

$2= \dfrac{0.5}{0.2535}+ \dfrac{0.5^2}{0.1719} - t_o (initial)$

$2 = 3.4267 - t_o (initial) \\ \\ t_o(initial) = 3.4267 - 2 \\ \\ t_o(initial) = 1.4267 \ hr$

NOW;

$1.4267 = \dfrac{d_o}{0.2535} + \dfrac{d_o^2}{0.1719} \\ \\ 1.4267 = 3.9448 \ d_o + 5.8173 \ d_o^2 \\ \\ d_o^2 + 0.6781 \ d_o = 0.2453$

$d_o = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

$d_o = \dfrac{-(0.6781) \pm \sqrt{(0.6781)^2-4(1)(-0.245)}}{2(10)}$

$d_o = \dfrac{-(0.6781) \pm \sqrt{0.45981961+0.98}}{20}$

$d_o = \dfrac{-(0.6781) \pm \sqrt{1.43981961}}{20}$

$d_o = \dfrac{-(0.6781) + \sqrt{1.43981961}}{20} \ OR \ \dfrac{-(0.6781) - \sqrt{1.43981961}}{20}$

$d_o =0.02609 \ OR \ -0.0939$

Thus; since we will consider the positive sign, the initial thickness $d_o$ is ;

≅ 0.261 μm

3 0

3 years ago