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Ierofanga [76]
3 years ago
14

Calculate the number of sulfate ions in 41.65 g of aluminum sulfate.

Chemistry
1 answer:
scoray [572]3 years ago
8 0

The number of sulfate ions : 2.2 x 10²³

<h3>Further explanation</h3>

Reaction

Al₂(SO₄)₃⇒2Al³⁺+3SO₄²⁻

mol Al₂(SO₄)₃(MW=342,15 g/mol) :

\tt \dfrac{41.65}{342.15}=0.122

mol of sulfate ions (SO₄²⁻)

\tt 3\times 0.122=0.366

1 mol = 6.02 x 10²³ particles (ions, molecules, atoms), so for 0.366 mol :

\tt 0.366\times 6.02\times 10^{23}=2.20\times 10^{23}

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Similar cells with the same function, all working together for the same purpose, form the 2nd level of organization in an organi
inn [45]

Answer:

tissue

Explanation:

cell→tissue→organ→systems→organism

7 0
3 years ago
HELP ME, MARKED AS BRAINEST
ELEN [110]

Answer:

The answer is B

Explanation:

2Na + 2H2O = 2NaOH + H2

5 0
3 years ago
If 17. 6 g of hcl are used to produce a 14. 5 l solution, what is the ph of the solution?.
kati45 [8]

This problem is providing us with the mass of hydrochloric acid and the volume of solution and asks for the pH of the resulting solution, which turns out to be 1.477.

<h3>pH calculations</h3>

In chemistry, one can calculate the pH of a solution by firstly obtaining its molarity as the division of the moles of solute by the liters of solution, so in this case for HCl we have:

M=\frac{17.6g*\frac{1mol}{36.46g} }{14.5L} \\\\M=0.0333 M

Next, due to the fact that hydrochloric acid is a strong acid, we realize its concentration is nearly the same to the released hydrogen ions to the solution upon ionization. Thereby, the resulting pH is:

pH=-log(0.0333)\\\\pH=1.477

Which conserves as much decimals as significant figures in the molarity.

Learn more about pH calculations: brainly.com/question/1195974

3 0
2 years ago
The empirical formula of a compound is CH. At 200 degree C, 0.145 g of this compound occupies 97.2 mL at a pressure of 0.74 atm.
Vladimir [108]

Answer:

The molecular formula = C_{6}H_{6}

Explanation:

Given that:

Mass of compound, m = 0.145 g

Temperature = 200 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (200 + 273.15) K = 473.15 K

V = 97.2 mL = 0.0972 L

Pressure = 0.74 atm

Considering,  

n=\frac{m}{M}

Using ideal gas equation as:

PV=\frac{m}{M}RT

where,  

P is the pressure

V is the volume

m is the mass of the gas

M is the molar mass of the gas

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the values in the above equation as:-

0.74\times 0.0972=\frac{0.145}{M}\times 0.0821\times 473.15

M=78.31\ g/mol

The empirical formula is = CH

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12 + 1 = 13 g/mol

Molar mass = 78.31 g/mol

So,  

Molecular mass = n × Empirical mass

78.31 = n × 13

⇒ n ≅ 6

The molecular formula = C_{6}H_{6}

6 0
3 years ago
One of the radioactive isotopes used in medical treatment or analysis is chromium-51. The half-life of chromium-51 is 28 days. H
Goryan [66]

Answer : The time required for decay is, 84 days.

Explanation :

Half-life of chromium-51 = 28 days

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{28\text{ days}}

k=0.0248\text{ days}^{-1}

Now we have to calculate the time required for decay.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = time taken by sample = ?

a = let initial activity of the sample = 100

a - x = amount left after decay process  = 12.5

Now put all the given values in above equation, we get

t=\frac{2.303}{0.0248}\log\frac{100}{12.5}

t=83.9\text{ days}\approx 84\text{ days}

Therefore, the time required for decay is, 84 days.

7 0
3 years ago
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