How many grams of aluminum are needed to react completely with 1.2 mol of FeO?

What is the term for the amount of energy that is needed for a chemical reaction to occur?

cupoosta [38]

This would be the activation energy. This is usually in the form of heat, which allows the reaction to undergo some sort of transition. Many times, enzymes can be used as catalysts to lower the activation energy required for the reaction to take place.

3 0

3 years ago

Question 4, can you give me the rule for naming these?

Scorpion4ik [409]

3 ethyl, 4 methylheptane. The compound is named by first identifying the longest carbon chain in the structure. in this case the chain has seven carbon atoms thus the prefix hept-.

Next you identify the substituent groups attached to the long carbon chain and name them from the lowest value of the integer assigned to the carbon atoms from either side. From the right, the ethyl group is attached to carbon number 3 while from the left, the methyl group is attached to carbon number 4. We therefore start with the right and name the attached groups first, including the carbon atoms to which they are attached.

Then we also take into consideration the highest number of bonds between the carbon atoms which is one from the question. Thus the suffix -ane is added if a maximum of one bond, -ene,if two bonds and -yne if three bonds.

3 0

3 years ago

Is pcl3 covalent or ionic

WITCHER [35]

Covalent Bond.

To be specific, it is polar covalent bond. :)

5 0

2 years ago

A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal

goblinko [34]

Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂:

$1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC$

All of the hydrogens atoms in the compound ended up becoming H₂O, so the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O:

$0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH$

Because the compound is composed only by C, H and O, the mass of O in the compound can be calculated by substraction:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C

mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H

mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

8 0

2 years ago

If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated

Dovator [93]

Answer : The molar mass of the solute would be low.

Explanation :

Formula used for depression in freezing point is:

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{w_b}{M_b}\times w_a}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of water

i = Van't Hoff factor

$K_f$ = freezing point constant

m = molality

$w_b$ = mass of solute

$w_a$ = mass of solvent

$M_b$ = molar mass of solute

From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.

Hence, the molar mass of the solute would be low.

6 0

3 years ago