What is the current model of atom

Form the complementary solution to the homogeneous equation. yc(t) = Cl [ ] +C2 Construct the (simplest) particular olution of t

In-s [12.5K]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

6 0

3 years ago

Which of the following statements is TRUE about the loose rocks on the surface

Ivan

Answer:B

Explanation:

3 0

3 years ago

Read 2 more answers

When working with hazardous substances in the laboratory your partner splashes a small amount of Sodium Hydroxide onto their clo

alexira [117]

Answer:

a) After helping our partner, we should immediately report the incident to the lab manager or any person in charge of the emergencies occurring in the lab.

b) We should have a copy of the Material Safety Data Sheet to give to the responders. This is because the responder can identify what materials were being used by the person ans what other security measures need to be taken.

5 0

3 years ago

The two elements discovered by Marie Curie are _____ and _____. uranium curium radium polonium

denpristay [2]

Marie Curie discovered polonium and radium. Hope this helps:))) Please, mark Brainliest!

5 0

2 years ago

Read 2 more answers

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.