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2 years ago

What mass of excess reactant remains at the end of the reaction if 90.0 g of so2 are mixed with 100.0 g of o2? 2so2 + o2 → 2so3?

1 answer:

8 0

For the reaction 2SO2 + O2 -> 2 SO3, we first determine which is the excess reactant between SO2 and O2. We list down the molar mass of the reactants:

Molar mass of SO2 = 64.0638 g/mol

Molar mass of O2 = 32 g/mol

Using the stoichiometry of the reaction, we then calculate the amount of oxygen that will react with 90.0 g of SO2.

90.0 g SO2 x 1 mol SO2/64.0638 g x 1 mol O2/ 2 mol SO2 x 32 g O2/mol = 22.4776 g O2

Thus, we can conclude that O2 is the excess reactant while SO2 is the limiting reactant. Subtracting 22.4776 g O2 from the initial 100.0 g O2, we get 77.5224 g O2 left after the complete reaction of 90.0 g SO2.

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Calculate the mass in grams of each of the following amounts: 1.002 mol of chromium 4.08 x 10-8 mol of neon

Pepsi [2]

Answer:

$Mass_{chromium}=52.1\ g$

$Mass_{neon}=8.23\times 10^{-7}\ g$

Explanation:

Calculation of the mass of chromium as:-

Moles = 1.002 moles

Molar mass of chromium = 51.9961 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$1.002\ mol= \frac{Mass}{51.9961\ g/mol}$

$Mass_{chromium}=1.002\times 51.9961\ g = 52.1\ g$

Calculation of the mass of neon as:-

Moles = $4.08\times 10^{-8}$ moles

Molar mass of neon = 20.1797 g/mol

Thus,

$1.002\ mol= \frac{Mass}{20.1797\ g/mol}$

$Mass_{neon}=4.08\times 10^{-8}\times 20.1797\ g = 8.23\times 10^{-7}\ g$

6 0

2 years ago

A person throws a ball up into the air, and the ball falls back toward Earth. At which point would the kinetic energy be the low

dybincka [34]

It would be d bc when it’s at its lowest point it means not that much energy and that would be its lowest point which is D

6 0

2 years ago

Read 2 more answers

5. Gabi has plans with her friends to go to a concert on her birthday in 4 days. She is so excited that she wants to know how ma

drek231 [11]

Answer:

So she is very anxious because she has to wait 345600 seconds

Explanation:

60 second = 1 minute

60 minute = 1 hour

1 hour has 3600 seconds (60*60)

24 hour = 1 day

3600 second * 24 hours =

1 day has 86400 seconds so in four days

86400 * 4 = 345600

7 0

2 years ago

HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.

Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, $C_A$ = 6.0mol/dm³

volume of HA, $V_A$ = 25.0cm³, = 0.025dm³

Concentration of HB, $C_B$ = 3.0mol/dm³

volume of HB, $V_B$ = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

$C_iVi = C_fV_f$

where;

$C_i$ is initial concentration

$V_i$ is initial volume

$C_f$ is final concentration of the solution

$V_f$ is final volume of the solution

$C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3$

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0

2 years ago

trans-2-Butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm-1); however, IR spectroscopy is

spayn [35]

Answer:

The other signal that would indicate the presence of a C= C bond appears close to 3100 $cm^{-1}$ .

Explanation:

Bands that appear above 3000 $cm^{-1}$ are often unsaturation diagnoses suggest. The band at 3000- 3100 $cm^{-1}$ is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.

4 0

2 years ago