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2 years ago

How many moles are in 25 grams of HF

2 answers:

6 0

Suppose 25 L to be 25 kg water (almost, -0.2% at room temperature) then divide 25,000 g by the moalr mass of H2O being 2+16=18 g/mole you get = 1385 moles (because 1 mole = 18 grams)

sergey [27]2 years ago

5 0

Using this equation, we can take 25/(1.0 + 19) and find that it is equal to 1.25 moles.

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Enter the chemical formula of a binary molecular compound of hydrogen and a Group 4A element that can reasonably be expected to

irakobra [83]

Answer: GeH4 (Germanium(IV) Hydride)

Explanation:

A Binary molecular compound Hydrogen and a Group 4A element which is more more acidic than SiH4 in aqueous solution is GeH4.

The pKa of GeH4;

= 25

Whilst that of SiH4

= 35

The lesser the pKa the higher the Ka which means more acidic.

6 0

2 years ago

How many atoms are in 578g Na

borishaifa [10]

1.51 x 10²⁵atoms

Explanation:

Given parameters:

Mass of Na = 578g

Unknown:

Number of atoms = ?

Solution:

To find the number of atoms, we must first find the number of moles the given mass contains.

Number of moles = $\frac{mass}{molar mass}$

molar mass of Na = 23g

Number of moles = $\frac{578}{23}$ = 25.13moles

1 mole of a substance = 6.02 x 10²³atoms

25.13 mole of Na = 25.13 x 6.02 x 10²³atoms

This gives 1.51 x 10²⁵atoms of Na

Learn more:

Avogadro's constant brainly.com/question/2746374

#learnwithBrainly

5 0

3 years ago

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su

anastassius [24]

Answer:

$V=27992L=28.00m^3$

Explanation:

Hello,

In this case, the combustion of methane is shown below:

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

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

$n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4$

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3$

Best regards.

8 0

3 years ago

He particles are mostly free of each other influence is solid , liquid , gas

Simora [160]

the most free particles are gases, then liquids, solids

4 0

3 years ago

At constant temperature a bicycle tire pump contains 252mL of air at 995kPa pressure. The plunger of the pump is pushed down unt

Novay_Z [31]

Answer:

The new pressure of the pump is 26.05 atm or 2639.4 kPa

Explanation:

Step 1: Data given

Volume of the bicycle tire pump = 252 mL = 0.252 L

Pressure of air = 995 kPa = 9.81989 atm

The volume of the pump is reduced to 95.0 mL = 0.095 L

Step 2: Calculate the new pressure

V1*P1 = V2*P2

⇒with V1 = the initial volume of the bicycle tire pump = 0.252 L

⇒with P1 = the initial pressure of the pump = 9.81989 atm = 995 kPa

⇒with V2 = the reduced volume of the pump = 0.095 L

⇒with P2 = the new pressure = TO BE DETERMINED

0.252 L * 9.81989 atm = 0.095 L * P2

P2 = 26.05 atm

The new pressure is 26.05 atm

0.252 L * 995 = 0.095 L * P2

P2 = 2639.4 kPa

The new pressure of the pump is 26.05 atm or 2639.4 kPa

4 0

3 years ago

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