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Delvig [45]
3 years ago
7

How many liters of oxygen are required to react with 2.5 grams of magnesium?

Chemistry
1 answer:
emmasim [6.3K]3 years ago
6 0

Answer:

1.16L

Explanation:

First, let us generate the balanced equation for the reaction. This is illustrated below

2Mg + O2 —> 2MgO

Now let us covert 2.5g of Mg given in the question to moles. This is illustrated below:

Molar Mass of Mg = 12g/mol

Mass of Mg = 2.5g

Number of mole of Mg =?

Number of mole = Mass /Molar Mass

Number of mole of Mg = 2.5/24 = 0.104mole

From the equation,

2moles of Mg required 1mole of O2.

Therefore, 0.104mole of Mg will require = 0.104/2 = 0.052mole of O2

Now let us calculate the volume occupy by 0.052mole of O2. This is shown below.

1mole of a gas occupy 22.4L at stp

Therefore, 0.052mole of O2 will occupy = 0.052 x 22.4 = 1.16L

Therefore, 1.16L of O2 is required for the reaction

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In a two-step synthesis, C6H11Br is converted into C6H12O. From the structure of the product, molecular formula of the starting
tamaranim1 [39]

Answer:

See explanation below

Explanation:

The question is incomplete. However in picture 1, you have the starting materials and the structure of the product, which you miss in this part.

Now, in picture 2, you have the starting reactant and the product, and the mechanism that is taking place here.

First, all what we have here is an acid  base reaction. In the first step, we are using the acid medium to convert the reactant into an alcohol. The bromine there, is not leaving the molecule yet, because it's neccesary for the next step. The starting reactant is an alkene, in that way, we can convert the reactant in the first step into a secondary alcohol. In other words, the first reaction is a alkene hydration.

In the second step, we use a strong base. You may say this is a strong nucleophile and will do a Sn2 reaction to form another alcohol there, but it's not the case, because, before any kind of reaction happens, the priority here is always the acid base, so the base will react with the acidic hydrogen. In this case, it will substract an hydrogen from the OH. When this happens, the lone pair will do an auto condensation here, and attacks the bromine in the molecule. In this way, the molecule will become a cyclomolecule, and that way it form the final product.

See picture 2, for mechanism

8 0
3 years ago
2
son4ous [18]

Answer: The molar enthalpy change is 73.04 kJ/mol

Explanation:

HCl+NaOH\rightarrow NaCl+H_2O

moles of HCl= molarity\times {\text {vol in L}}=0.415mol/L\times 0.1=0.0415mol

As NaOH is in excess 0.0415 moles of HCl reacts with 0.0415 moles of NaOH.

volume of water = 100.0 ml + 50.0 ml = 150.0 ml

density of water = 1.0 g/ml

mass of water = volume \times density=150.0ml\times 1.0g/ml=150.0g

q=m\times c\times \Delta T

q = heat released

m = mass  = 150.0 g

c = specific heat = 4.184J/g^0C

\Delta T = change in temperature = 4.83^0C

q=150.0\times 4.184\times 4.83

q=3031.3J

Thus 0.0415 mol of HCl produces heat = 3031.3 J

1 mol of HCL produces heat = \frac{3031.3}{0.0415}\times 1=73043.3J=73.04kJ

Thus molar enthalpy change is 73.04 kJ/mol

8 0
3 years ago
You have 2.7 moles of carbon. How many atoms do you have?
Gelneren [198K]

Answer:

1.63 × 10²⁴ atoms.

Explanation:

To calculate the number of atoms (N) contained in 2.7moles of carbon, we multiply the number of moles (n) by Avogadro's number (6.02 × 10²³).

That is, N = n × nA

Where;

N = number of atoms

n = number of moles (mol)

nA = Avogadro's numbe

N = 2.7 × 6.02 × 10²³

N = 16.254 × 10²³

N = 1.63 × 10²⁴ atoms.

Hence, there are 1.63 × 10²⁴ atoms in 2.7moles of Carbon.

4 0
3 years ago
If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of
Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.1\ g}{186.04\ g/mol}

Moles\ of\ ferrocene= 0.0005375\ mol

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

\rho=\frac{Mass}{Volume}

or,  

Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.044\ g}{78.49\ g/mol}

Moles\ of\ acetyl\ chloride= 0.0005606\ mol

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms  0.0005375 mole of monoacetylferrocene

Moles of product formed =  0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>

5 0
3 years ago
12. Complete each of the following word equations for synthesis reactions.
Vlada [557]

Explanation:

12)

a) Sodium + oxygen = ?

When sodium reacts oxygen its forms sodium oxide as a product.

4Na(s)+O_2(g)\rightarrow 2Na_2O(s)

b) magnesium + fluorine = ?

When magnesium reacts with fluorine its forms magnesium fluoride as a product.

Mg(s)+F_2(g)\rightarrow MgF_2(s)

13)

a) 2HgO(s)\overset{\Delta }\rightarrow 2Hg(l)+O_2(g)

Decomposition of mercury(II) oxide on heating gives out mercury and oxygen gas.

b)2H_2O(l)\overset{Electricity}\rightarrow 2H_2(g)+O_2(g)

The electrolytic decomposition of water gives out hydrogen gas amd oxygen gas.

5 0
3 years ago
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