What is the term called ¨I nutted

What is the total mass of 2.0 moles of diatomic hydrogen?

sladkih [1.3K]

diatomic hydrogen is written as H2 (2.02 grams H2) <------- if each hydrogen atom is 1.01 grams, then two hydrogen atoms are 2.02 grams 2.0 moles H2 X 2.02 grams H2 ------------- (divide to cancel moles) = 4.04 grams/mole H2 ÷ one mole = 4.04 grams H2

7 0

3 years ago

Read 2 more answers

Acetic acid, ethanol, acetaldehyde and ethane form a series of 2-carbon molecules which differ in their extent of oxidation. Con

yanalaym [24]

Explanation:

Ethanol can be oxidized to ethanal or acetaldehyde which is further oxidized to acid that is acetic acid.

$CH_3CH_2OH$ → $CH_3CHO$ [oxidation by loss of hydrogen]

An oxidizing agent potassium dichromate(VI) solution is used to remove the hydrogen from the ethanol.

An oxidizing agent used along with dilute sulphuric acid for acidification.

Acetaldehyde can also be reduced back to ethanol again by adding hydrogen to it by using a reducing agent that is sodium tetrahydro borate, NaBH4.

The oxidation of aldehydes to carboxylic acids can be done by the two-step process.

In the first step, one molecule of water is added in the presence of a catalyst that is acidic.

There is a generation of a hydrate. (geminal 1,1-diol).

$CH_3CHO$ → $CH_3CH_2COOH$ [reduction by the gain of electrons]

Here, the oxidizing agent used is $Cr_3O$ in the presence of acetone.

7 0

3 years ago

Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this

maw [93]

Answer: d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

$2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H=+1411.1kJ$

Reversing the reaction, changes the sign of $\Delta H$

$C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$

$\Delta H=-1411.1kJ$

On multiplying the reaction by $\frac{1}{2}$ , enthalpy gets half:

$0.5C_2H_4(g)+1.5O_2(g)\rightarrow CO_2(g)+H_2O(l)$ $\Delta H=\frac{1}{2}\times -1411.1kJ=-705.55kJ/mol$

Thus the enthalpy change for the given reaction is -705.55kJ

7 0

3 years ago

Answer these please ASAP need help no idea how to do these

STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass = g

Cl₂:

Number of moles = Mass / molar masa

Number of moles = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol = Mass / 2 g/mol

Mass = 16 g

P₄:

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles = 1.6 g /48 g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol = mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO → 2Fe + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

Fe : F₂O₃

2 : 1

1.78 : 1/2×1.78 = 0.89 mol

Mass of F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂ → 2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

Al : AlI₃

2 : 2

0.05 : 0.05

I₂ : AlI₃

3 : 2

0.1 : 2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂ → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

Pb : O₂

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass = 0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂ → 2PbO

6 0

3 years ago

Please I need help on these 3 questions. Thank You.

Liula [17]

1.

V = 200 mL (volume)

c = 3 M = 3 mol/L (concentration)

First we convert mL to L:

200 mL = 0.2 L

Then we calculate the moles using the formula: n = V × c = 0.2 L × 3 mol = 0.6 mol

Finally, we just use the molar mass of CaF2 to calculate the actual mass:

molar mass = 78 g/mol

The formula is: m = n × mm (mass = moles × molar mass)

m = 0.6 mol × 78 g/mol = 46.8 g

2.

For this question the steps are exactly like the first question.

V = 50mL = 0.05 L

c = 12 M = 12 mol/L

n = V × c = 0.05 L × 12 mol/L = 0.6 mol

molar mass (HCl) = 36.5 g/mol

m = n × mm = 0.6 mol × 36.5 g/mol = 21.9 g.

3.

The steps for this question are the opposite way.

m(K2CO3) = 250 g

molar mass = 138 g/mol

n = m ÷ mm = 1.81 mol

c = 2 mol/L

V = n ÷ c = 1.81 mol ÷ 2 mol/L = 0.905 L = 905 mL

6 0

3 years ago