An unknown liquid has a mass of 30.6 g

Compare and contrast single-bonded compounds and double-bonded compounds.

max2010maxim [7]

The main difference between single double and triple bond is the number of shared electrons. If the shared number is one pair of electrons, the bond will be a single bond, whereas if two atoms bonded by two pairs (four electrons), it will form a double bond.

Souce: pediaa.com/difference-between-single-double-and-triple-bonds/

3 0

4 years ago

a gas cylinder contains 2.0 mol of gas X and 6.0 mol of gas Y at a total pressute of 2.1 atm. what is he partial pressure of gas

Igoryamba

The answer is 1.6 atm. Let's first calculate the mole fraction of gas Y.

The mole fraction (x) is: x = n1/n, where n1 is a number of moles of an

individual gas in a mixture and n is total moles of the gas mixture. We

know that n1 of gas Y is 6.0 mol (n1 = 6.0 mol) and that there are in

total 8. mol of the gas mixture (n = 2.0 + 6.0 = 8.0 mol). Now calculate

the mole fraction of gas Y. x = 6.0/8.0 = 0.75. Now, let's use the mole

fraction of gas Y (x) and the total pressure (P) to calculate the

partial pressure of gas Y (P1): x = P1/P. P1 = x * P. If x = 0.75 and P =

2.1 atm, then the partial pressure of gas Y is: P1 = 0.75 * 2.1 atm =

1.6 atm.

6 0

4 years ago

Read 2 more answers

What's the equation for the reaction of Trifluoro bromide with Ammonia gas. ( brainliest assured )

sp2606 [1]

Answer:

Ammonium bromide can be prepared by the direct action of hydrogen bromide on ammonia. It can also be prepared by the reaction of ammonia with iron(II) bromide or iron(III) bromide, which may be obtained by passing aqueous bromine solution over iron filings.

Explanation:

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5 0

3 years ago

A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution

Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

<u>Step 1:</u> Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

<u>Step 2:</u> Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

<u>Step 3:</u> Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

<u>Step 4:</u> Calculate change in internal energy

ΔU = Q + W W = 0 (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

4 0

4 years ago

Which type of scientist would most likely study how electricity could be

Nikolay [14]

Electrical engineers are the scientists that study the production of electricity more efficient.

<h3>Which type of scientist would study how electricity could be produced more efficiently?</h3>

Electrical engineers are the type of scientist that would most likely study how electricity could be produced more efficiently because they produced new technologies which are cost efficient as well as produce high amount of electricity.

So we can conclude that electrical engineers are the scientists that study the production of electricity more efficient.

Learn more about electricity here: brainly.com/question/776932

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6 0

2 years ago