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3 years ago

How many hydrogen atom are present in 1.53 g of water

Chemistry

1 answer:

svp [43]3 years ago

3 0

Answer:

$=1.02x10^{23} atoms\ H$

Explanation:

Hello,

In this case, for water, whose molar mass is 18 g/mol, we can find two moles of hydrogen in one mole of water, therefore, for us compute the atoms, we should also use the Avogadro's number as shown below:

$=1.53gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O} *\frac{6.022x10^{23}atoms\ H}{1molH} \\\\=1.02x10^{23} atoms\ H$

Regards.

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Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It decomposes by the reaction: COCl

Oksana_A [137]

Answer:

[CO] = 7.61x10⁻³M

7.61x10⁻³x10³ = 7.61

Explanation:

For a generic equation aA + bB ⇄ cC + dD, the constant of equilibrium (Kc) is:

$Kc = \frac{[C]^cx[D]^d}{[A]^ax[B]^b}$

We need to know the molar concentrations in the equilibrium. In the beginning, there is only COCl₂, and its concentration is the number of moles divided by the volume:

[COCl₂] = 7.73/10.0 = 0.773 M

So, the equilibrium will be:

COCl₂(g) ⇆ CO(g) + Cl₂(g)

0.773 0 0 <em>Initial</em>

-x +x +x <em> Reacts</em>

0.773-x x x <em>Equilibrium</em>

Supposing that x<<0.773, then:

$Kc = \frac{x*x}{0.773}$

7.5x10⁻⁵ = x²/0.773

x² = 5.7975x10⁻⁵

x = √5.7975x10⁻⁵

x = 7.61x10⁻³ M

The supposing is correct, so [CO] = 7.61x10⁻³ x 10³ = 7.61

6 0

3 years ago

Calcular la normalidad de 1 Kg de sulfuro de aluminio en 5000 ml de solucion.

Troyanec [42]

Calculate the normality of 1 Kg of aluminum sulfide in 5000 ml of solution.

Normality comes out to be 8.11

<h3> Given </h3>

Mass of solute: 1000g
Volume of solution (V): 5000 ml = 5 liters
Equivalent mass of solute (E) = molar mass / n-factor

n-factor for $Al_{2}S_{3}$ is 6 and molar mass is 148g

So, on calculating equivalent mass is equal to 24.66g

FORMULAE of Normality (N) = (Mass of the solute) / (Equivalent mass of the solute (E) × Volume of the solution (V)

N= $\frac{1000}{24.66*5}$

Therefore, normality of 1 kg aluminum sulfide is 8.11

Learn more about normality here brainly.com/question/25507216

#SPJ10

7 0

1 year ago

Chlorine forms a number of oxides with the following oxidation numbers: 1, 3, 4, 6, and 7. These compounds are stable and neutra

SashulF [63]

Answer:

$Cl_2O$ , $Cl_2O_3$ , $ClO_2$ , $ClO_3$ , $Cl_2O_7$

Explanation:

Empirical formula of the compound is the simplest ratio of elements present in the compound.

Empirical formula of compounds of chlorine with oxygen is as follows:

Compounds in which oxidation state of Cl is +1

$Cl_2O$

Compounds in which oxidation state of Cl is +3

$Cl_2O_3$

Compounds in which oxidation state of Cl is +4

$ClO_2$

Compounds in which oxidation state of Cl is +6

$ClO_3$

Compounds in which oxidation state of Cl is +7

$Cl_2O_7$

8 0

3 years ago

What is the oxidation state of S in HSO4^2-? (A.P.E.X)

BlackZzzverrR [31]

I think there is a typo because I've never seen HSO4 2- before in my life. It should be HSO4-. For that, H is 1+ and each Oxygen is 2-0 totaling 8-. So the oxidation state of sulfur +1 - 8 = 7

So the oxidation state of sulfur is +6

7 0

2 years ago

For the following reaction, 35.4 grams of zinc oxide are allowed to react with 6.96 grams of water . zinc oxide(s) + water(l) --

IRISSAK [1]

Answer:

$m_{Zn(OH)_2}=38.4g$

Explanation:

Hello!

In this case, for the undergoing chemical reaction:

$ZnO(s)+H_2O(l)\rightarrow Zn(OH)_2$

We evaluate the yielded moles of zinc hydroxide by each reactant as shown below:

$n_{Zn(OH)_2}^{by ZnO}=35.4gZnO*\frac{1molZnO}{81.38gZnO}*\frac{1molZn(OH)_2}{1molZnO} =0.435molZn(OH)_2\\\\n_{Zn(OH)_2}^{by H_2O}=6.96gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{1molZn(OH)_2}{1molH_2O} =0.386molZn(OH)_2$

In such a way, since the water yields a smaller amount of zinc hydroxide we conclude it is the limiting reactant so the maximum mass is computed below:

$m_{Zn(OH)_2}=0.386molZn(OH)_2*\frac{99.424 gZn(OH)_2}{1molZn(OH)_2} \\\\m_{Zn(OH)_2}=38.4g$

Because the water limits the yielded amount of zinc hydroxide.

Best regards!

5 0

3 years ago