I need help with this question please Chemistry

How many moles in 4.3 x 1024 atoms of sodium (Na)?

wel

Answer:How many ions are in 2 moles of NaCl?

2 mol NaCl = 2 x 6.02 x 10 23 = 1.204 x 10 24 NaCl formula units 2 mol NaCl = 2 (2 x 6.02 x 10 23 )= 2 (1.204 x 10 24) = 2.408 x 10 24 atoms *2 moles of sodium chloride contains 2.408 x 10 24 ions (combination of Na + ions & Cl – ions). H

Explanation:

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3 0

2 years ago

A 41.1 g sample of solid CO2 (dry ice) is added to a container at a temperature of 100 K with a volume of 3.4 L.A. If the contai

marta [7]

Answer:

Approximately 6.81 × 10⁵ Pa.

Assumption: carbon dioxide behaves like an ideal gas.

Explanation:

Look up the relative atomic mass of carbon and oxygen on a modern periodic table:

C: 12.011;
O: 15.999.

Calculate the molar mass of carbon dioxide $\rm CO_2$ :

$M\!\left(\mathrm{CO_2}\right) = 12.011 + 2\times 15.999 = 44.009\; \rm g \cdot mol^{-1}$ .

Find the number of moles of molecules in that $41.1\;\rm g$ sample of $\rm CO_2$ :

$n = \dfrac{m}{M} = \dfrac{41.1}{44.009} \approx 0.933900\; \rm mol$ .

If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:

$P \cdot V = n \cdot R \cdot T$ ,

where

$P$ is the pressure inside the container.
$V$ is the volume of the container.
$n$ is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
$R$ is the ideal gas constant.
$T$ is the absolute temperature of the gas.

Rearrange the equation to find an expression for $P$ , the pressure inside the container.

$\displaystyle P = \frac{n \cdot R \cdot T}{V}$ .

Look up the ideal gas constant in the appropriate units.

$R = 8.314 \times 10^3\; \rm L \cdot Pa \cdot K^{-1} \cdot mol^{-1}$ .

Evaluate the expression for $P$ :

$\begin{aligned} P &=\rm \frac{0.933900\; mol \times 8.314 \times 10^3 \; L \cdot Pa \cdot K^{-1} \cdot mol^{-1} \times 298\; K}{3.4\; L} \cr &\approx \rm 6.81\times 10^5\; Pa \end{aligned}$ .

Apply dimensional analysis to verify the unit of pressure.

4 0

3 years ago

Why does stirring increase the rate of dissolution?

gavmur [86]

The correct option is D.

When dissolving a substance in a solvent, stirring the solution will increased the rate at which the substance dissolved. This is because, when one stirs a solution, it exposes more surface area of the solute to the solvent, thus, increasing the interaction between the solute and the solvent. The higher the quantity of the solute that is exposed to the solvent, the higher the rate of dissolution of the solute.

5 0

2 years ago

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A sample of pure alumina hydrate was obtained. A 5.000 g sample of the material was heated carefully in a vacuum oven until no m

lapo4ka [179]

Answer: The formula of the hydrated alumina is $Al_2O_3.5H_2O$

Explanation:

Decomposition of hydrated alumina is given by:

$Al_2O_3.xH_2O\rightarrow Al_2O_3+xH_2O
$

Molar mass of $Al_2O_3$ = 101.96 g/mol

According to stoichiometry:

(101.96+18x) g of $Al_2O_3.xH_2O$ decomposes to give 101.96 g of

$Al_2O_3$

Thus 5.000 g of $Al_2O_3.xH_2O$ decomposes to give= $\frac {101.96}{(101.96+18x)}\times 5.000$ of H_2O

But it is given 5.000 g of a sample of hydrated salt $Al_2O_3.xH_2O$ was found to contain 2.6763 g of unhydrated salt

Thus we can equate the two equations:

$\frac{101.96}{(101.96+18x)}\times 5.000=2.6763$

$x=5$

Thus the formula of the hydrated alumina is $Al_2O_3.5H_2O$

6 0

2 years ago

Iodine is a black, crystalline, shiny solid and belongs to Group VII. Sodium is a soft, shiny solid and belongs to Group I.

skad [1K]

<h2>Answer -A </h2>

As a pure element, iodine is a lustrous purple-black nonmetal that is solid under standard conditions. It sublimes (changes from a solid to a gaseous state while bypassing a liquid form) easily and gives off a purple vapor. Although it is technically a non-metal, it exhibits some metallic qualities.

<h2>Answer - B</h2>

Broadly speaking, there are four types of classification. They are: (i) Geographical classification, (ii) Chronological classification, (iii) Qualitative classification, and (iv) Quantitative classification

3 0

2 years ago

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