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3 years ago

What is the correct way to show carbon dioxide in the gas phase?

1 answer:

8 0

The correct way of showing carbon dioxide in the gas phase is b. CO2 (g)

This notation is commonly used in showing chemical reactions wherein the phases of the constituents are important. Gases are denoted by (g), solids are denoted by (s), and liquids are denoted by (l). Aqueous solutions are also denoted (aq).

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DDT was banned from use as a pesticide in the United States because

Dmitry_Shevchenko [17]

Too much money and dangerous

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3 years ago

Arun and Marcus are asked to compare the solubility of two salts in water at room temperature. The two salts are labelled X and

oksian1 [2.3K]

Answer:

1) Salts X and Y

2) The solubility of the salts

3) a) The solvent

b) The solvent temperature

Explanation:

1) The independent variable is the variable that is suspected to be the cause of the subject of the investigation

The given investigation is meant to investigate the solubility of different salts

Therefore, the solubility is expected to be dependent on the type of salt, and the independent variable is the type of salt, X or Y

2) The dependent variable is the effect meant to be observed in the investigation, which is the solubility of the salt in water at room temperature

3) The control variables are the variables which are held constant during the investigation, including;

a) The solvent used if the investigation; water

b) The temperature of the solvent; Room temperature

3 0

3 years ago

13. How many moles of zinc are in 25.00 g Zn?

topjm [15]

0.3824 moles of Zinc

4 0

3 years ago

Suppose that 7.25 x 10^22 atoms of a hypothetical element have a mass of 3.88 g. What would be the molar mass (g/mol) of this el

S_A_V [24]

Answer:

32.23 to 4 significant figures.

Explanation:

The molar mass of the element is the mass of 6.022 * 10^23 atoms (Avogadro's number).

So by proportion it is 6.022 * 10^23 * 3.88 / 7.25 * 10^22

= 32.23 to 4 significant figures.

4 0

3 years ago

A 0.8870 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 t

bagirrra123 [75]

Answer : The percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess $AgNO_3$ then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.8870 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

$\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles$

$\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.8870-x)g}{74.5g/mole}=\frac{(0.8870-x)}{74.5}moles$

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = $\frac{x}{58.5}moles$

Moles of chloride ions in KCl = $\frac{(0.8870-x)}{74.5}moles$

The total moles of chloride ions = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

$\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{2.142g}{143.32g/mole}=0.0149moles$

Now we have to determine the value of 'x'.

Moles of AgCl = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

0.0149 mole = $\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles$

By solving the term, we get the value of 'x'.

$x=0.818g$

The mass of NaCl = x = 0.818 g

The mass of KCl = (0.8870 - x) = 0.8870 - 0.818 = 0.069 g

Now we have to calculate the mass percent of NaCl and KCl.

$\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.818g}{0.8870g}\times 100=92.22\%$

$\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.069g}{0.8870g}\times 100=7.78\%$

Therefore, the percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.

3 0

3 years ago