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3 years ago

Using the standardized NaOH solution in the previous

Chemistry

2 answers:

geniusboy [140]3 years ago

6 0

Answer:

Using the standardized NaOH solution in the previous question, the titration of 5.00 mL of vinegar required 42.25 mL of NaOH. What is the % of acetic acid in the vinegar? Concentration of NaOH from previous question= 0.0986 M. David C

Explanation:

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Studentka2010 [4]3 years ago

5 0

Answer: Using the standardized NaOH solution in the previous question, the titration of 5.00 mL of vinegar required 42.25 mL of NaOH. What is the % of acetic acid in the vinegar? Concentration of NaOH from previous question= 0.0986 M. David C.

Explanation:

Please mark as brainliest for me.Thank you so much

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Is chromium a transition metal

oksano4ka [1.4K]

No it is not

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7 0

3 years ago

Read 2 more answers

One of the intermediates in the synthesis of glycine from ammonia, carbon dioxide, and methane is aminoacetonitrile, C2H4N2. The

sleet_krkn [62]

Answer:

Mass of C₂H₄N₂ produced = 3.64 g

Explanation:

The balanced chemical equation for the reaction is given below:

3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)

From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O

Molar masses of the compounds are given below below:

CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol

Comparing the mole ratios of the reacting masses;

CH₄ = 1.65/16 = 0.103

CO₂ = 13.5/44 = 0.307

NH₃ = 2.21/17 = 0.130

converting to whole number ratios by dividing with the smallest ratio

CH₄ = 0.103/0.103 = 1

CO₂ = 0.307/0.103 = 3

NH₃ = 0.130/0.103 = 1.3

Multiplying through with 5

CH₄ = 1 × 5 = 5

CO₂ = 3 × 5 = 15

NH₃ = 1.3 × 5 = 6.5

Therefore, the limiting reactant is NH₃

8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂

Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂

Mass of C₂H₄N₂ produced = 3.64 g

7 0

3 years ago

An FM radio station broadcast at a frequency of 98.5 MHz. What is the wavelength of the station’s broadcast signal?

Troyanec [42]

Answer:

(1 MHz = 106 s -1)

Explanation:

4 0

3 years ago

Hydrochloric acid reacts with sodium hydroxide to produce sodium chloride and water. If 20.6 g of sodium hydroxide reacts with a

astraxan [27]

Answer:

30.1 g NaCl

Explanation:

Your first conversion is converting grams NaOH to moles of NaOH using its molar mass (39.997 g/mol). Then, use the mole ratio of 1 mol NaCl for every 1 mol NaOH to get to moles of NaCl. Then finally multiply by the molar mass of NaCl (58.44 g/mol) to get grams of NaCl.

20.6 g NaOH • (1 mol NaOH / 39.997 g NaOH) • (1 mol NaCl / 1 mol NaOH) • (58.44 g NaCl / 1 mol NaCl) = 30.1 g NaCl

4 0

3 years ago

consider the reaction between calcium oxide and carbon dioxide: cao ( s ) + co 2 ( g ) → caco 3 ( s ) a chemist allows 14.4 g of

MakcuM [25]

Answer:

CaO is the limiting reagent

Theoritical yield = 25.71 g

% Yield = 75.44%

Explanation:

1 mole = Molar mass of the substance

Molar Mass of CaO = 56 g/mol

Molar Mass of CaCO3 = 100 g/mol

Molar mass of CO2 = 44 g/mol

The balanced Equation is :

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO reacts with = 1 mole of CO2

56 g of CaO reacts with = 44 g of CO2

1 g of CaO reacts with =

$\frac{44}{56}$

= 0.785 g of CO2

So,

14.4 g of CaO must react with = (14.4 x 0.785) g of CO2

= 11.31 g of CO2

Needed = 11.31 g

Available CO2 = 13.8 g (given)

So CO2 is in excess , hence CaO is the limiting reagent and product will produce from 14.4 g of CaO

$CaO + CO_{2}\rightarrow CaCO_{3}$

1 mole of CaO will produce 1 mole pf CaCO3

56 g of CaO produce = 100 g of CaCO3

1 g of CaO produce =

$\frac{100}{56}$

= 1.785 g of CaCO3

14.4 g of CaO will produce = (1.785 x 14.4) g of CaCO3

= 25.71 g of CaCO3

Theoritical Yield of CaCO3 = 25.71 g

Actual yield = 19.4 g

Percent Yield =

$\frac{Actual\ yield}{Theoritical\ yield}\times 100$

$\frac{19.4}{25.71}\times 100$

= 75.44 %

6 0

3 years ago