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2 years ago

Which has the larger radius, F or F− ?

1 answer:

6 0

Answer is B- F has a smaller radius than F− because an additional electron causes greater repulsion in F− is the correct choice and the nuclear charge the radius of the anion increases.

Explanation: I hoped that helped!

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If 2.0 liters of oxygen gas react with excess carbon monoxide at 273 K and 1.00 atm, how many liters of carbon dioxide can be pr

vlada-n [284]

The given condition is STP, under this condition, gas has a rule of 22.4 L per mole. And the given equation is already balanced. The ratio of mole number is the same as the ratio of the volume and is also the same as the ratio of coefficients. So the answer is 4.0 liters.

5 0

2 years ago

A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.

dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

3 0

2 years ago

Investigate and explain that when a force is applied to an object but it does not move, it is because another opposing force is

N76 [4]

Answer:

4. Force 1 Force 2 Comparison of Forces

Friction Gravity Force 1 is equal to Force 2

Explanation:

5 0

2 years ago

Read 2 more answers

Describe how you would make 6.0 L of a 0.100 M KCl solution from a 4.0 M stock KCl solution.

Galina-37 [17]

Answer:

0.15 l of 4.0 m stock KCl solution should betaken

Explanation:

N1V1=N2V2

6*0.1=V2*4

V2=0.15L

5 0

2 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

2 years ago