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2 years ago

Which molecule is an alkene?

Chemistry

1 answer:

belka [17]2 years ago

4 0

D is an alkene because the ending is cis-2-pentene if it was an alkane is would be pentane.

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If the equilibrium concentrations of products are much greater than those of reactants in this system, what would be the magnitu

miss Akunina [59]

Answer:

K > 1.

Explanation:

∵ The equilibrium constant K = [products]/[reactants].

Since, [products] > [reactants].

∴ The equilibrium constant K > 1.

5 0

3 years ago

Which of the following is most likely to lose electrons in an ionic compound?

ANTONII [103]

Answer:

a. Lead (Pb)

Am 100% sure

Hope it helps

please mark me as brainliest

4 0

2 years ago

A chemist must prepare 800.0mL of sodium hydroxide solution with a pH of 12.10 at 25°C. She will do this in three steps: Fill a

AnnZ [28]

Answer:

0.42 g

Explanation:

We have:

pH = 12.10 (25 °C)

V = 800.0 mL = 0.800 L

To find the mass of sodium hydroxide (NaOH) we can use the pH:

$14 = pH + pOH$

$pOH = 14 - pH = 14 - 12.10 = 1.90$

$pOH = -log ([OH^{-}])$

$[OH]^{-} = 10^{-pOH} = 10^{-1.90} = 0.013 M$

Now, we can find the number of moles (η) of OH:

$\eta = ([OH]^{-})*V = 0.013 mol/L * 0.800 L = 1.04 \cdot 10^{-2} moles$

Since we have 1 mol of OH in 1 mol of NaOH, the number of moles of NaOH is equal to 1.04x10⁻² moles.

Finally, with the number of moles we can find the mass of NaOH:

$m = \eta * M$

Where M is the molar mass of NaOH = 39.9 g/mol

$m = 1.04 \cdot 10^{-2} moles * 39.9 g/mol = 0.42 g$

Therefore, the mass of sodium hydroxide that the chemist must weigh out in the second step is 0.42 g.

I hope it helps you!

3 0

3 years ago

Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction entropy of the following chemical rea

Lapatulllka [165]

Answer:

-122 J/K

Explanation:

Let's consider the following balanced reaction.

N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g)

We can calculate the standard reaction entropy (ΔS°) using the following expression.

ΔS° = Σ ηp × Sf°p - Σ ηr × Sf°r

where,

η: stoichiometric coefficients of products and reactants

Sf°r: entropies of formation of products and reactants

ΔS° = 2 mol × 240.06 J/K.mol - 1 mol × 191.61 J/K.mol - 2 mol × 205.14 J/K.mol

ΔS° = -121.77 J/K ≈ -122 J/K

8 0

2 years ago

How many moles of argon gas would be present in a 37.0 liter vessel at 45.00 °C at a pressure of 2.50 atm?

Pani-rosa [81]

Answer:

3.54 mol

Explanation:

Step 1: Given data

Volume (V): 37.0 L

Temperature (T): 45.00 °C

Pressure (P): 2.50 atm

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 45.00°C + 273.15 = 318.15 K

Step 3: Calculate the number of moles (n) of argon gas

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 2.50 atm × 37.0 L/(0.0821 atm.L/mol.K) × 318.15 K = 3.54 mol

8 0

3 years ago