If the highest point on this landform is 650 feet and the lowest point is 0 feet, what is the relief of this landform?

Write the electron configuration for N and N-3. How many electrons are in the valence energy level?

Alinara [238K]

Non-valence electrons: 1s22s22p6. Therefore, we write the electron configuration for Na: 1s22s22p63s1. What is the highest principal quantum number that you see in sodium's electron configuration? It's n = 3, so all electrons with n = 3 are valence electrons, and all electrons with n < 3 are non-valence electrons.

8 0

3 years ago

Isoflavones, quercetin and anthocyanins are all what type of phytochemicals

kykrilka [37]

Isoflavones, quercetin, and anthocyanin are all Flavonoids.

8 0

3 years ago

Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o

stich3 [128]

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of butane

$\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles$

b) moles of oxygen

$\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles$

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

According to stoichiometry :

2 moles of butane require 13 moles of $O_2$

Thus 0.09 moles of butane will require = $\frac{13}{2}\times 0.09=0.585moles$ of $O_2$

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

7 0

3 years ago

What other objects are often arranged by there properties

Trava [24]

Answer:

other than the periodic table, are often arranged by their properties

Explanation:

Money, they are arranged by how big or small they are, like coins, with pennies, dimes, and quarters.

4 0

3 years ago

Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib

KengaRu [80]

Answer:

For methanol: Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol: Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol: Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

For methanol:

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

For ethanol:

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

For propanol:

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

$moles=\frac{Mass(m)}{Molar\ mass (M)}$

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

5 0

3 years ago