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nordsb [41]
3 years ago
11

If 1.20 moles of an ideal gas occupy a volume of 18.2 l at a pressure of 1.80 atm, what is the temperature of the gas, in degree

s celsius?
Chemistry
1 answer:
scoundrel [369]3 years ago
6 0

We can calculate for temperature by assuming the equation for ideal gas law:

P V = n R T

Where,

P = pressure = 1.80 atm

V = volume = 18.2 L

n = number of moles = 1.20 moles

R = gas constant = 0.08205746 L atm / mol K

Substituting to the given equation:

T = P V / n R

T = (1.8 atm * 18.2 L) / (1.2 moles * 0.08205746 L atm / mol K)

T = 332.70 K

We can convert K unit to ˚C unit by subtracting 273.15 to Kelvin, therefore

T = 59.55 ˚<span>C</span>

You might be interested in
1 Ammonia, NH3, reacts with incredibly strong bases to produce the amide ion, NH2 -. Ammonia can also react with acids to produc
Charra [1.4K]

Answer:

a) ammonium ion

b) amide ion

Explanation:

The order of decreasing bond angles of the three nitrogen species; ammonium ion, ammonia and amide ion is NH4+ >NH3> NH2-. Next we need to rationalize this order of decreasing bond angles from the valence shell electron pair repulsion (VSEPR) theory perspective.

First we must realize that all three nitrogen species contain a central sp3 hybridized carbon atom. This means that a tetrahedral geometry is ideally expected. Recall that the presence of lone pairs distorts molecular structures from the expected geometry based on VSEPR theory.

The amide ion contains two lone pairs of electrons. Remember that the presence of lone pairs causes greater repulsion than bond pairs on the outermost shell of the central atom. Hence, the amide ion has the least H-N-H bond angle of about 105°.

The ammonia molecule contains one lone pair, the repulsion caused by one lone pair is definitely bless than that caused by two lone pairs of electrons hence the bond angle of the H-N-H bond in ammonia is 107°.

The ammonium ion contains four bond pairs and no lone pair of electrons on the outermost nitrogen atom. Hence we expect a perfect tetrahedron with bond angle of 109°.

5 0
3 years ago
Someone pls help me I will make you brain
Katena32 [7]

Answer:

it is actually b because i did this i picked b and got it right

Explanation:

4 0
3 years ago
C. What is the mathematical equation for the total pressure below the
faltersainse [42]
I can answer your question if you have a picture or a writing, your welcome.
8 0
3 years ago
Read 2 more answers
When 28.0 g of acetylene reacts with hydrogen, 24.5 g of ethane is produced. What is the percent yield of C2H6 for the reaction?
Afina-wow [57]

Answer:

Y=75.6\%

Explanation:

Hello.

In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:

m_{C_2H_6}^{theoretical}=28.0gC_2H_2*\frac{1molC_2H_2}{26gC_2H_2}*\frac{1molC_2H_6}{1molC_2H_2}  *\frac{30gC_2H_6}{1molC_2H_6}\\ \\m_{C_2H_6}^{theoretical}=32.3gC_2H_6

Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:

Y=\frac{24.5g}{32.3g}*100\%\\ \\Y=75.6\%

Best regards.

3 0
3 years ago
Can someone help me with 2 and 3? I don't know the answer and I need justifications for why the correct answers are correct
saw5 [17]

Answer:

2. (C) K < 1.

3. (B) [Fe³⁺] = 2.00 mol·L⁻¹; [SCN⁻] = 6.0 mol·L⁻¹

Step-by-step explanation:

2. Value of K

A⇌ B

K = [B]/[A]

If the concentration of reactants (A) is larger than the concentration of products (B), the denominator of the K expression is larger than the numerator.

The fraction is less than 1, so

K < 1

3. Equilibrium concentrations

We can use an ICE table to keep track of the calculations.

                     Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺

I/mol·L⁻¹:     6.00      10.0            0

C/mol·L⁻¹:     -x           -x             +x

E/mol·L⁻¹:  6.00-x  10.0-x            x

Initially, there is no FeSCN²⁺ present, so [FeSCN²⁺] = 0.

Then, the Fe³⁺ and SCN⁻ react until equilibrium is reached.

How much will react? We don't know, but we have every confidence that x mol (some unknown quantity) will react.

[Fe³⁺] will <em>decrease</em> by x mol·L⁻¹. Because of the 1:1:1 molar ratios, [SCN⁻] will also <em>decrease</em> by x mol·L⁻¹ and [FeSCN²⁺] will <em>increase</em> by x mol·L⁻¹.

We add the changes and get the values in the bottom line.

However, what is the value of that pesky x?

We are told that [FeSCN²⁺] = 4.00 mol·L⁻¹ at equilibrium.

From the table, x = [FeSCN²⁺], so x = 4.00.

Now we can insert these values back into the table.

At equilibrium,

[Fe³⁺]   =   6.00 - x =  6.00 - 4.00 = 2.00 mol·L⁻¹

[SCN⁻] = 10.0    - x = 10.0   - 4.00 = 6.0   mol·L⁻¹

5 0
3 years ago
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