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Wittaler [7]
3 years ago
13

For the balanced equation shown below, if the reaction of 77.0 grams of CaCN2 produces 26.1 grams of NH3, what is the percent yi

eld?
CaCN2+3H2O=>CaCO3+2NH3
Chemistry
1 answer:
11111nata11111 [884]3 years ago
4 0

%yield = 86.93%

<h3>Further explanation</h3>

Given

Reaction

CaCN₂+3H₂O=>CaCO₃+2NH₃

Required

The percent yield

Solution

mol CaCN₂(MW= 92 g/mol) :

= mass : MW

= 77 g : 92 g/mol

= 0.834

From equation, mol NH₃ :

= 2/1 x mol CaCN₂

= 2/1 x 0.834

= 1.668

Mass NH₃(theoretical):

= 1.668 mol x 18 g/mol

= 30.024 g

% yield = (actual/theoretical) x 100%

%yield = (26.1/30.024) x 100%

<em>%yield = 86.93%</em>

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The two naturally occurring isotopes of antimony, 121Sb (57.21 percent) and 123Sb (42.79 percent), have masses of 120.904 and 12
Alex

Answer:

The correct answer is option c.

Explanation:

Formula used to determine an average atomic mass :

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

Mass of isotope Sb-121 = 120.904 amu

Fractional abundance of Sb-121 = 57.21% = 0.5721

Mass of isotope Sb-123 = 122.904 amu

Fractional abundance of Sb-123 = 42.79% = 0.4279

Average atomic mass of Sb:

120.904 amu\times 0.5721+ 122.904 amu\times 0.4279=121.7598 amu \approx 121.76 amu

7 0
3 years ago
You are given sodium acetate, 1m hcl, nahco3 and na2co3. Determine which of these four you would need and then show calculations
valentina_108 [34]

solution:

the given compounds are sodium acetate, 1M HCL,NaHCO₃ and Na2CO₃

pH of the buffer solution is 4.7

the value of pKa of sodium bicarbonate is 6.37

the value of pKa of acetic acid is 4.7

calculate concentration of acetic acid by using the following forumula

pH=pKa+lag[salt]/[acid]

substitute the pH and Pka values in the formula.

4.7=4.7+log[salt]/[acid]

log[salt]/[acid]=0

thus, the concentration ratio of the salt and acid should be equal to each other.

Thus, concentration of sodium acetate is 0.05M

Concentration of sodium acetate= concentration of acid

= 0.05M

Volume of the buffer solution is 100mL

The buffer solution can be prepared as 0.05M of 50mL sodium acetate will react with 0.05M of 50mL of 0.05M of HCL.

The chemical equation for neutralization of the weak base with strong can be represented as show as

CH₃COONa+HCL-->CH₃COOH+NaCL


5 0
3 years ago
What is the penetrating power of beta
zlopas [31]
The penetrating power of alpha rays, beta rays, and gamma rays varies greatly. Alpha particles can be blocked by a few pieces of paper. Beta particles pass through paper but are stopped by aluminum foil. Gamma rays are the most difficult to stop and require concrete, lead, or other heavy shielding to block them.
3 0
3 years ago
When a 0.245-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.260-g sample
sveticcg [70]

Answer:

The heat of combustion per moles of caffeine is 4220 kJ/mol

Explanation:

Step 1: Data given

⇒ When  benzoic acid sample of 0.245 grams is burned the temperature rise is 1.643 °C

⇒ When 0.260 gram of caffeine is burned, the temperature rise is 1.436 °C

⇒ Heat of combustion of benzoic acid = 26.38 kJ/g

<u>Step 2:</u> Calculate the heat released: for combustion of benzoic acid

0.245 g benzoic acid *  26.38 kJ/g = 6.4631 kJ

<u>Step 3</u>: Calculate the heat capacity of the calorimeter:

c = Q/ΔT

Q = 6.4631 kJ   / 1.643°C = 3.934 kJ/ °C

<u>Step 4:</u> Calculate moles of a 0.260 g sample of caffeine:

Moles caffeine = Mass caffeine / Molar mass caffeine

0.260 grams/ 194.19 g/mol  = 0.0013389 moles

Step 5: Calculate heat released: for combustion of caffeine

Q = c * ΔT

Q = 3.934 kJ/°C * 1.436 °C = 5.65 kJ

Step 6: Calculate the heat of combustion per mole of caffeine  

5.65 kJ  /  0.0013389 moles = 4219.9 kJ/mol  ≈ 4220 kJ/mol

The heat of combustion per moles of caffeine is 4220 kJ/mol

4 0
4 years ago
I Need help with this worksheet I Need the Work Please Thank you
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1. a. 3

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4 0
2 years ago
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