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tekilochka [14]

2 years ago

7

Given that cao(s) + h2o(l) → ca(oh)2(s), δh°rxn = –64.8 kj/mol, how many grams of cao must react in order to liberate 525 kj of

heat?

1 answer:

USPshnik [31]2 years ago

8 0

Answer:

454.3 g.

Explanation:

From the given data:

1.0 mol of CaO liberates → – 64.8 kJ.

??? mol of CaO liberates → - 525 kJ.

∴ The no. of moles needed = (1.0 mol)(- 525 kJ)/(- 64.8 kJ) = 8.1 mol.

<em>∴ The no. of grams of CaO needed = no. of moles x molar mass</em> = (8.1 mol)(56.077 g/mol) = <em>454.3 g.</em>

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The correct name for hno2 is ________. nitrous acid hyponitrous acid nitric acid pernitric acid hydrogen nitrate

balu736 [363]

Nitrous Acid.

Hyponitrous acid: H2N2O2

Nitric acid: HNO3

Pernitric acid: HNO

4 0

2 years ago

A sample of krypton has a volume of 6.00 L, and the pressure is 0.960 atm. If the final temperature is 55.0°C, the final volume

MissTica

193.38 K was the initial temperature of the krypton.

Explanation:

Data given:

Initial volume of the krypton gas = 6 litres

initial pressure of the krypton gas = 0.960 atm

initial temperature of the krypton gas = ?

final volume of the krypton gas = 7.70 litres

final pressure of the Krypton gas = 1.25 atm

final temperature of the krypton gas = 55 degrees or 273.25+55 = 323.15 K

Applying the Combined Gas Laws:

$\frac{P1V1}{T1} = \frac{P2V2}{T2}$

Rearranging the equation:

T1 = $\frac{P1V1T2}{P2V2}$

Putting the value in the equation:

T1 = $\frac{0.960 X 6 X 323.15}{1.25 X 7.70}$

T1 = 193.38 K

Initial temperature of the krypton gas is 193.78 K

5 0

2 years ago

A sample of a radioactive isotope initially contains 20 x 10^10 atoms. After 16 days, 5 x 10^10 atoms remain. what is the half-l

nekit [7.7K]

I think the answer is A but I could be wrong

7 0

2 years ago

Can 750 mL of water dissolve 0.60 mol of gold(III) chloride (AuCl3)?

nydimaria [60]

Answer:

Yes

Explanation:

1. Mass of 0.60 mol of AuCl₃

$\text{Mass} = \text{0.60 mol} \times \dfrac{\text{303.33 g}}{\text{1 mol}} = \text{184 g}$

2. Mass of AuCl₃ in 750 mL

The solubility of AuCl₃ is 68 g/100 mL.

In 750 mL of water, you can dissolve

$\text{Mass of AuCl}_{3} = \text{750 mL} \times \dfrac{\text{68 g}}{\text{100 mL}} = \text{510 g AlCl}_{3}$

∴ Yes, 750 mL of water can dissolve 0.60 mol of AuCl₃.

3 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

Read 2 more answers