Please someone help I’m really confused

Is the color of the flam a test for the nitrate polyatomic ion in each of the compounds?

Valentin [98]

4 0

3 years ago

What is the concentration of each ion in a solution that is prepared by dissolving 5.00 g of ammonium chloride in enough water t

sukhopar [10]

The concentration of each ion in the solution of ammonium chloride is:

NH₄⁺ = 0.1886 M
Cl⁻ = 0.1886 M

To solve this problem, the formulas and the procedures that we have to use are:

M = n(solute)/v(solution) L
n = m / MW
MW= ∑ AWT

Where:

M= molarity
n = moles
m = mass
v = volume
MW = molecular weight
AWT = atomic weight

Information about the problem:

m(NH₄Cl) = 5 g
v(solution) = 500 ml
AWT (N)= 14 g/mol
AWT (H)= 1 g/mol
AWT (Cl) = 35 g/mol

Converting the volume units from (ml) to (L) we have:

v(solution) = 500 ml * (1 L/1000 ml)

v(solution) = 0,50 L

We calculate the moles of the NH₄Cl from the MW:

MW = ∑ AWT

MW (NH₄Cl)= AWT (N) + AWT (H)*4 + AWT (Cl)

MW (NH₄Cl)= 14 g/mol +( 1 g/mol * 4) + 35 g/mol

MW (NH₄Cl)= 14 g/mol + 4 g/mol + 35 g/mol

MW (NH₄Cl)= 53 g/mol

Having the MW we calculate the moles of NH₄Cl:

n(NH₄Cl) = m(NH₄Cl) / MW(NH₄Cl)

n(NH₄Cl) = m(H2SO4) / MW (H2SO4)

n(NH₄Cl) = 5 g / 53 g/mol

n(NH₄Cl) = 0.0943 mol

Applying the molarity formula, we get:

M(NH₄Cl) = n(NH₄Cl)/v(solution) L

M(NH₄Cl) = 0.0943 mol / 0,50 L

M(NH₄Cl) = 0.1886 M

There are 0.1886 moles of NH₄Cl per liter of solution.

Let's recognize that 1 mol NH₄Cl contains:

1 mol NH₄⁺
1 mol Cl⁻

The concentration of each ion is thus:

(1)*(0.1886 M) = 0.1886 M NH₄⁺

(1)*(0.1886 M) = 0.1886 M Cl⁻

<h3>What is a solution?</h3>

In chemistry a solution is known as a homogeneous mixture of two or more components called:

Solvent
Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

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4 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

3 years ago

Read 2 more answers

How many atoms are in a sample of 68.7 g copper (Cu)?

dlinn [17]

<span>Divide the number of grams present in the sample by copper's gram atomic weight to find the number of gram atomic weights present. Then multiply that result by Avogadro's Number: 6.022137 x 10^23 atoms/gram atomic weight.1,200 g/(63.54 g/gram atomic weight) ? 18.885741 gram-atomic weights. Hope this helps. </span>

8 0

4 years ago

Read 2 more answers

What is the correct order of increasing density?

klasskru [66]

Answer:

(Density = mass/volume) Arrange the following in order of increasing density - air, exhaust from chimneys, honey, water, chalk, cotton and iron. Answer: The increasing order of density is air < exhaust from chimney < cotton < water < honey < chalk < iron.

Explanation:

7 0

3 years ago