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ziro4ka [17]
3 years ago
6

How does a balanced chemical equation satisfy the law of conservation of mass

Chemistry
1 answer:
ELEN [110]3 years ago
4 0
Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. ... When an equal number of atoms of an element is present on both sides of a chemical equation, the equation is balanced.
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2.20 moles Sn to grams
mihalych1998 [28]
261.162 grams. Use the equation n=M÷Mr.
3 0
3 years ago
Water is a source of ______________ for chemical reactions in cells.
S_A_V [24]

Answer:Electrolyte

Explanation:Electrolyte is a compound either in solution or in molten form that gets decomposed by thee passage if electricity through it. Water is polar,but a weak electrolyte, so in cells,water is used as an electrolyte

5 0
3 years ago
A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa
Liula [17]

Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

4 0
3 years ago
Many esters have characteristic odors. Methyl salicylate exhibits a pleasant wintergreen odor. This ester can be prepared from s
hodyreva [135]

Answer:

Methanol would be used as a reagent in excess, since it is a very low-cost solvent. For product isolation, the first thing to do is remove the methanol through a distillation process. The residue produced can be dissolved in diethyl ether. Using a NaHCO₃ solution, extraction is performed. When it separates into two phases, the product will be in the ether and the reagent in the aqueous phase. The ether can also be removed by distillation, and at the end of this process you will have the product you want.

Explanation:

5 0
3 years ago
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
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