Lanthanide: tungsten chromium promethium

Chemistry

2 answers:

Lisa [10]4 years ago

8 0

The answer on Edgenuity2020 is promethium

grandymaker [24]4 years ago

4 0

Answer:

promethium

Explanation:

edge2020

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The weight percent of concentrated H2SO4, molar mass=980 g/mol, is 960% and its density is 184 g/ml. What is the molarity of con

Elan Coil [88]

Answer:

<em>C</em> H2SO4 = 9.79 M

Explanation:

molarity (M) ≡ # dissolved species / V sln
H2SO4 ↔ H3O+ + SO4-

∴ %w/w H2SO4 = 960% = g H2SO4 / g sln * 100

⇒ 9.6 = g H2SO4 / g sln

calculation base: 1000 g sln

⇒ g H2SO4 = 9600g

⇒<em> </em>mol<em> </em>H2SO4 =<em> </em>9600 g H2SO4 * ( mol H2SO4/ 980g H2SO4 ) = 9.796 mol H2SO4

⇒ V sln = 1000g sln / 1000g/L = 1 L sln

∴ ρ H20 ≅ 1000 Kg/m³ = 1000 g/L

⇒ <em>C</em> H2SO4 = 9.796 mol H2SO4 / 1 L sln

⇒ <em>C</em> H2SO4 = 9.796 M

8 0

4 years ago

How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethy

malfutka [58]

Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

Explanation:

First we have to calculate the moles of diethyl ether

$\text{Moles of diethyl ether}=\frac{\text{Mass of diethyl ether}}{\text{Molar mass of diethyl ether}}=\frac{75.0g}{74g/mole}=1.01moles$

As, 1 mole of diethyl ether require heat = 26.5 kJ

So, 1.01 moles of diethyl ether require heat = $\frac{26.5}{1}\times 1.01=26.8kJ$

Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

5 0

4 years ago

How many moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas?

tino4ka555 [31]

Taking into account the reaction stoichiometry, 340.0 moles of methane are produced when 85.1 moles of carbon dioxide gas react with excess hydrogen gas

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CO₂ + 4 H₄ → CH₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CO₂: 1 mole
H₄: 4 moles
CH₄: 1 mole
H₂O: 2 moles

<h3>Moles of CH₄ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of CO₂ form 4 moles of CH₄, 85.1 moles of CO₂ form how many moles of CH₄?

$moles of CH_{4} =\frac{85.1 moles of CO_{2}x4 moles of CH_{4} }{1 moles of CO_{2}}$