Which formula id correctly paired with its name? A. MgSO↓ = magnesium chlorine

OlgaM077 [116]

I think the answer is yes

3 0

2 years ago

A mixture of helium, nitrogen and oxygen has a total pressure of 821 mmHg. The partial pressure of helium is 105 mmHg, and the p

skad [1K]

Answer:

Total partial pressure, Pt = 821 mm Hg

Partial pressure of Helium, P1 = 105 mm Hg

Partial pressure of Nitrogen, P2 = 312 mm Hg

Partial pressure of Oxygen, P3 = ? mm Hg

According to Dalton's law of Partial pressures,

Pt = P1 + P2 + P3

So, <u>P3 = 404 mm Hg</u>

4 0

2 years ago

Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: $Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S$

First, balance the chemical equation

$Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S$

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

$n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol$

From the chemical reaction , the ratio of molar is 3mol $H_{2}S/1 mol Al_{2}S_{3}.$ So, the moles of hydrogen sulfide are:

$n_{H_{2} O} =\frac{131g}{18.02g/mol}$

= 7.26mol

From the chemical reaction, the molar ratio is 3 mol $H_{2}S/6$ mol $H_{2}O.$ So, the moles of hydrogen sulfide are:

Moles of $H_{2}S$ formed = 7.26 mol $H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }$

Th liming reactant is $Al_{2}S_{3}$ beacuse the mass of $Al_{2}S_{3}$ forms less product than water. Therefore, the maximum number of moles of $H_{2}S$ is 3.15 mol. We know that molar mass of $H_{2}S$ is 34.10g/mol. So, the maximum mass of $H_{2}S$ formed is,

$m_{H_{2}S } = n_{H_{2}S } \times$ Molar mass of $H_{2}S$

= 3.15 mol $\times$ 34.10g/mol

= 107.4g

Now, multiplying the number of moles of $Al_{2}S_{3}$ by the molar ratio between $Al_2S_3$ and $H_2O$ which is 6mol $H_2O/1mol$ $Al_2S_3$ we get the number of moles of $H_2O$ reacted.

Moles of $H_2O$ reacted = 1.05 mol $Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}$

= 6.31 mol $H_2O$

The mass of $H_2O$ is,

$m_{H_{2} O}$ = 6.31 mol $\times$ 18.02g/ mol

= 114g

On subtracting, the mass of $H_2O$ reacted from the given mass of $H_2O$ is,

$m_{H_2O}$ = (131-114)g

= 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

#SPJ4

8 0

2 years ago

Determine how many gmol, kmol, and lbmols there are in 50 kilograms of n-hexane.

Artist 52 [7]

Answer: 581 gmol

0.581 kmol

$1.28\times 10^{-3}lbmol$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{50\times 1000g}{86g/mol}=581mol$

1. The conversion for mol to gmol

1 mol = 1 gmol

581 mol= $\frac{1}{1}\times 581=581gmol$

2. The conversion for mol to kmol

1 mol = 0.001 kmol

581 mol= $\frac{0.001}{1}\times 581=0.581kmol$

3. The conversion for mol to lbmol

1 mol = $2.2\times 10^{-3}lbmol$

581 mol= $\frac{2.2\times 10^{-3}}{1}\times 581=1.28\times 10^{-3}lbmol$

3 0

3 years ago

Which of the following elements is the most metallic

Marysya12 [62]

We found cesium, strontium, aluminum, sulfur, chlorine, and fluorine on the periodic table. Cesium is the farthest left and the lowest, while fluorine is the farthest right and the highest, so we know they have the highest metallic character and the lowest metallic character, respectively.

7 0

3 years ago

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