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3 years ago

Although tom was a poor student, he studied very well

1 answer:

3 0

Tom. The ‘T’ should be in capital. Every sentence should end with a full stop ‘.’ . This also has nothing to do with Chemistry. So yeah. Mark me brainliest please.

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Plete the following problems significant figures: 6 × 0.30<br>

jeka57 [31]

6= Only the digits 1 and 6 are the actual measured values. Therefore we have only 2 significant figures.

0.3= Zeros used as placeholders are not significant. Zeros that come before non-zero integers are never significant. Example 5: The zeros in 098, 0.3, and 0.000000000389 are not significant because they are all in front of non-zero integers. c. If the zeros come after non-zero integers and are followed by a decimal point, the zeros are significant.

3 0

3 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

El NaCI corresponde a un elemento?

Schach [20]

Answer:

Sodio (Na), elemento químico del grupo de los metales alcalinos (Grupo 1 [Ia]) de la tabla periódica. El sodio es un metal blanco plateado muy suave. El sodio es el metal alcalino más común y el sexto elemento más abundante en la Tierra, y comprende el 2,8 por ciento de la corteza terrestre.

Explanation:

Espero que esto ayude a marcar el MÁS CEREBRAL !!!

7 0

3 years ago

What is the pressure in atmospheres of the gas remaining in the flask? Ignore the volume of solid NH4Cl produced by the reaction

Mashcka [7]

Answer:

a) HCl is the limiting reagent.

b) Mass of NH₄Cl formed = 6.68 g

c) Pressure of the gas remaining in the flask = 1.742 atm

Explanation:

The complete Question is presented in the attached image to this solution.

To solve this question, we first need to obtain the limiting regaent for this reaction.

The limiting reagent is the reagent that is in short supply in the reaction and is used up in the reaction. It determines the amount of products that will be formed and the amount of other reactants that will be required for the reaction.

NH₃ (g) + HCl (g) ⟶ NH₄Cl (s)

1 mole of NH₃ reacts with 1 mole of HCl

we first convert the masses of the gases available to number of moles.

Number of moles = (Mass/Molar Mass)

Molar mass of NH₃ = 17.031 g/mol, Molar mass of HCl = 36.46 g/mol

Number of moles of NH₃ = (4.55/17.031) = 0.2672 mole

Number of moles of HCl = (4.55/36.46) = 0.1248 mole

Since 1 mole of NH₃ reacts with 1 mole of HCl

It is evident that HCl is in short supply and is the limiting reagent.

NH₃ is in excess.

So, to calculate the amount of NH₄Cl formed,

1 mole of HCl gives 1 mole of NH₄Cl

0.1248 mole of HCl will also gove 0.1248 mole of NH₄Cl

Mass (Number of moles) × (Molar Mass)

Molar mass of NH₄Cl = 53.491 g/mol

Mass of NH₄Cl formed = 0.1248 × 53.491 = 6.68 g

c) The gas remaining in the flask is NH₃

0.1248 mole of NH₃ is used up for the reaction, but 0.2672 mole was initially available for reaction,

The amount of NH₃ left in the reacting flask is then

0.2672 - 0.1248 = 0.1424 mole.

Using the ideal gas Equation, PV = nRT

We can obtain the rrequired pressure of the remaining gas in the flask

P = Pressure = ?

V = Volume = 2.00 L

n = number of moles = 0.1424 mole

R = molar gas constant = 0.08205 L.atm/mol.K

T = absolute temperature in Kelvin = 25 + 273.15 = 298.15 K

P = (nRT/V)

P = (0.1424×0.08205×298.15/2) = 1.742 atm

Hope this Helps!!!

7 0

4 years ago

What does GCAT help us remember?

vazorg [7]

It helps us remember the four bases of dna

7 0

3 years ago