Which of the following is NOT an example of a physical change? *

What is the molarity of a solution containing 23 g of NaCl in 500 mL of solution?

trapecia [35]

molarity of a solution means mols per liter.

First, you need to convert 23 grams on NaCl into mols. 23g divided by molar mass (58.44g/mol) which gives you .394 mols.

Now, you need to convert 500ml to L which moves the decimal three places to the left, giving you .500L of solution.

Finally, divide the mols over solution to get .787M

7 0

2 years ago

A slice of pizza has 500 kcal. If we could burn the pizza and use all the heat to warm a 50-L container of cold water, what woul

gayaneshka [121]

Answer:

10°C

Explanation:

Heat gain by water = Heat lost by the slice of pizza

Thus,

$m_{water}\times C_{water}\times \Delta T=Q$

<u>For water: </u>

Volume = 50.0 L

Density of water= 1 kg/L

So, mass of the water:

$Mass\ of\ water=Density \times {Volume\ of\ water}$

$Mass\ of\ water=1 kg/L \times {50.0\ L}$

Mass of water = 50 kg

Specific heat of water = 1 kcal/kg°C

ΔT = ?

For slice of pizza:

Q = 500 kcal

So,

$50\times 1\times \Delta T=500$

ΔT = 10°C

Increase in temperature = 10°C

3 0

2 years ago

Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium

Eddi Din [679]

Explanation:

It is known that $pK_{a}$ value of acetic acid is 4.74. And, relation between pH and $pK_{a}$ is as follows.

pH = pK_{a} + log $\frac{[CH_{3}COOH]}{[CH_{3}COONa]}$

= 4.74 + log $\frac{1.00}{1.00}$

So, number of moles of NaOH = Volume × Molarity

= 71.0 ml × 0.760 M

= 0.05396 mol

Also, moles of $CH_{3}COOH$ = moles of $CH_{3}COONa$

= Molarity × Volume

= 1.00 M × 1.00 L

= 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

$CH_{3}COONa + NaOH \rightarrow CH_{3}COOH$

Initial : 1.00 mol 1.00 mol

NaoH addition: 0.05396 mol

Equilibrium : (1 - 0.05396 mol) 0 (1.00 + 0.05396 mol)

= 0.94604 mol = 1.05396 mol

As, pH = pK_{a} + log $\frac{[CH_{3}COONa]}{[CH_{3}COOH]}$

= 4.74 + log $\frac{0.94604}{1.05396}$

= 4.69

Therefore, change in pH will be calculated as follows.

pH = 4.74 - 4.69

= 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

8 0

2 years ago

9

marishachu [46]

Answer:

B. as a food preservative in the manufacture of detergents

.........

8 0

2 years ago

The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

2 years ago

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