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3 years ago

The substances present before a reaction are called?

Chemistry

1 answer:

aniked [119]3 years ago

3 0

Answer:

( 1 )The substances present before a reaction are called? Products

( 2 ) In a chemical equation, reactant(s) and products are separated by what? An arrow.

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Each 5-ml teaspoon of Extra Strength Maalox Plus contains 450 mg of magnesium hydroxide and 500 mg of aluminum hydroxide. How ma

Art [367]

Answer:

0.0347 moles of hydronium ions

Explanation:

The equation of the neutralization reaction between hydroxide and hydronium ions is given below:

H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)

From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.

The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:

Number of moles = mass / molar mass

Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol

Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol

Mass of magnesium hydroxide = 450 g = 0.45 g

Mass of aluminium hydroxide = 500 mg = 0.5 g

Moles of magnesium hydroxide = (0.45/58) moles

Moles of aluminium hydroxide = (0.5/78) moles

Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:

Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)

Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)

Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles

Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles

Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions

Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.

3 0

3 years ago

Derek needs a tool that delivers 25.00 mL of a sodium hydroxide solution. What tool would be best for him to use?

3241004551 [841]

He may use a pipette with a volume greater than 25 ml to transfer the solution.

4 0

4 years ago

How many grams of h2 will be produced if 175g of HCI are allowed to react completely with sodium

Sedbober [7]

Answer:

4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium

Explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

HCl: 2 moles
Na: 1 mole
NaCl: 2 moles
H₂: 1 mole

You know the following masses of each element:

H: 1 g/mole
Cl: 35.45 g/mole
Na: 23 g/mole

So, the molar mass of each compound participating in the reaction is:

HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
Na: 23 g/mole
NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
H₂: 2* 1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:

HCl: 2 moles* 36.45 g/mole= 72.9 g
Na: 1 mole* 23 g/mole= 23 g
NaCl: 2 moles* 58.45 g/mole= 116.9 g
H₂: 1 mole* 2 g/mole= 2 g

So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

$mass of H_{2} =\frac{175 g of HCl*2g ofH_{2} }{72.9 g of HCl}$

mass of H₂= 4.8 g

<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>

3 0

3 years ago

Read 2 more answers

The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

Irina-Kira [14]

Answer:

$t=147.24\ min$

Explanation:

Given that:

Half life = 30 min

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30}\ min^{-1}$

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

$750e^{-0.0231t}=25$

$x=\frac{\ln \left(30\right)}{0.0231}$

$t=147.24\ min$

6 0

4 years ago

Each energy sublevel contains __________ number of electrons. For example, sublevel D can hold up to _______ electrons. A. the s

Annette [7]

Answer:

Each energy sublevel contains a different number of electrons. For example, sublevel D can contain up to 10 electrons

Explanation:

The atoms are surrounded by propellers that within each propeller there is a certain number of electrons, these electrons jump from orbit to orbit according to the amount of energy they have. The four levels that make up the electronic cloud that surrounds an atom are: s p d f.

When these electrons change orbit or level they release energy in the form of light, which is known as a photon.

3 0

3 years ago