Which number has four significant figures? 0.05730 meters 40.007 meter 0.00073 meter 8765 meter

Due to the fact that ionic compounds have strong intermolecular forces they are __________ at room temperature.

MrMuchimi

The correct answer should be solids

3 0

3 years ago

One crystalline form of silica (SiO2) has a cubic unit cell, and from X-ray diffraction data it is known that the cell edge leng

tigry1 [53]

Answer:

There are 8Si atoms and 16 O atoms per unit cell

Explanation:

From the question we are told that:

Edge length $l=0.700nm=>0.7*10^9nm$

Density $\rho=2.32g/cm^3$

Generally the equation for Volume is mathematically given by

$V=l^3$

$V=(0.7*10^9)m^3$

$V=3,43*10^-{22}cm$

Where

Molar mass of (SiO2) for one formula unit

$M=28+32$

$M=60g/mol$

Therefore

Density of Si per unit length is

$\rho_{si}=\frac{9.96*10^{23}}{3.43*10^22}$

$\rho=0.29$

Molar mass of (SiO2) for one formula unit

$M=28+32$

$M=60g/mol$

Therefore

There are 8Si atoms and 16 O atoms per unit cell

6 0

3 years ago

A dilute solution of bromine in carbon tetrachloride behaves as an ideal-dilute solution. The vapour pressure of pure CCl4 is 33

ANEK [815]

Explanation:

The given data is as follows.

Vapour pressure of pure $CCl_{4}$ = 33.85 Torr

Temperature = 298 K

Mole fraction of $Br_{2}$ = 122.36 torr

Therefore, calculate the vapor pressure of $Br_{2}$ as follows.

Vapour pressure of $Br_{2}$ = mole fraction of $Br_{2}$ x K of $Br_{2}$

= 0.050 x 122.36 Torr

= 6.118 Torr

So, vapor pressure of $Br_{2}$ is 6.118 Torr .

Now, calculate the vapor pressure of carbon tetrachloride as follows.

Vapour pressure of $CCl_{4}$ = mole fraction of $CCl_{4}$ x Pressure of $CCl_{4}$

= (1 - 0.050) × 33.85 Torr

= 32.1575 Torr

So, vapor pressure of $CCl_{4}$ is 32.1575 Torr .

Hence, the total pressure will be as follows.

= 6.118 Torr + 32.1575 Torr

= 38.2755 Torr

Therefore, composition of $CCl_{4}$ = $\frac{32.1575 Torr}{38.2755 Torr}$

= 0.8405

Composition of $CCl_{4}$ is 0.8405 .

And, composition of $Br_{2}$ = $\frac{6.118 Torr}{38.2755 Torr}$

= 0.1598

Composition of $Br_{2}$ is 0.1598 .

6 0

3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago

The chemical agent or hazardous material that interferes with the body's ability to transfer oxygen to the cells is:

zheka24 [161]

Phosgene its a my friend used in the war

7 0

3 years ago

Read 2 more answers