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QUESTION ABOUT IONIC COMPUND DUE IN 7 MINUTES URGENT!!!!!!!!!!!!!

nekit [7.7K]

Answer:

NH2S4. . ...............................,..

4 0

3 years ago

Take a paperclip into water and then slowly pull it upward to the point where it is nearly free from the surface. You'll find th

Mandarinka [93]

Answer:

Adhesive forces and cohesive forces

Explanation:

Cohesive forces is how much the water molecules stick to each other and adhesive forces are to which extinct they stick to other surfaces.

When we pull the paperclip upward the water molecules adhere to it due to adhesive forces. While the water on the clip is remain connected to water in the glass because of cohesive forces.

So if the adhesive forces dominates the water will be pulled out and if the cohesive forces dominate it will remain with other water in the cup.

5 0

3 years ago

Calculate the volume of carbon dioxide at 20.0°C and 0.941 atm produced from the complete combustion of 4.00 kg of methane. Comp

tankabanditka [31]

Answer:

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

Explanation:

Methane

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Mass of methane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of methane = $\frac{4000 g}{16 g/mol}=250 mol$

According to reaction, 1 mole of methane gives 1 mole of carbon dioxide gas,then 250 moles of methane will give :

$\frac{1}{1}\times 250 mol=250 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 250 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{250 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6390.89 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6390.89 Liters.

Propane

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Mass of propane = 4.00 kg = 4000 g (1 kg = 1000 g)

Moles of propane = $\frac{4000 g}{44 g/mol}=90.91 mol$

According to reaction, 1 mole of propane gives 3 mole of carbon dioxide gas,then 90.91 moles of propane will give :

$\frac{3}{1}\times 90.91 mol=272.73 mol$ of carbon dioxide gas

Moles of carbon dioxide gas = n = 272.73 mol

Pressure of carbon dioxide gas = P = 0.941 atm

Temperature of carbon dioxide gas = T = 20.0°C = 20.0+273 K = 293 K

Volume of carbon dioxide gas = V

$PV=nRT$ (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{272.73 mol\times 0.0821 atm L/mol K\times 293 K}{0.941 atm}=6,971.95 L$

The volume of carbon dioxide at 20.0°C and 0.941 atm produced was 6,971.95 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of methane = 6390.89 Liters.

Volume of carbon dioxide given by combustion of 4.00 kg of propane = 6,971.95 Liters.

6390.89 Liters < 6,971.95 Liters

More volume of carbon dioxide gas was released on combustion of 4.00 kg of propane in comparison to the volume of carbon dioxide released on combustion of 4.00 kg of methane.

7 0

3 years ago

Given: 4.39m/sec Go: km/hr

monitta

Answer:

1 m = 1/1000 km; 1 sec = 1/3600 hr

1 m/sec = 3600/1000 km/hr = 18/5 km/hr

Explanation:

3 0

3 years ago

1. Consider the chemical equation. If there are 40 mol of NBr3 and 48 mol

marusya05 [52]

Answer:

NaOH

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2NBr3 + 3NaOH —> N2 + 3NaBr + 3HOBr

The limiting reactant can be obtained as follow:

From the balanced equation above,

2 moles of NBr3 reacted with 3 moles of NaOH.

Therefore, 40 moles of NBr3 will react with = (40 x 3)/2 = 60 moles of NaOH.

From the above calculations,

A higher amount of NaOH i.e 60 moles than what was given from the question i.e 48 moles is required to react completely with 40 moles of NBr3. Therefore, NaOH is the limiting reactant and NBr3 is the excess reactant.

7 0

3 years ago

Need help please. bhhhj​

Need help please. bhhhj