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2 years ago

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You use an analytical balance to measure the mass of a weigh boat and record a mass of 1.5624 g. You then add some NaHCO3 to the

weigh boat and give the weigh boat to your lab partner to measure. Your lab partner mistakenly forgets to record all of the digits on the balance and records a mass of 1.92 g. What is the minimum number of grams of acetic acid that you would need to fully react with your NaHCO3

1 answer:

OverLord2011 [107]2 years ago

4 0

Answer:

0.2554g of acetic acid is the minimum mass you would need for a full reaction.

Explanation:

The minimum mass your partner could meadure is 1.9200 g. That means the mass of NaHCO₃ could be added is:

1.9200g - 1.5624g = 0.3576g of NaHCO₃ that reacts with acetic acid, (CH₃COOH), thus:

CH₃COOH + NaHCO₃ CH₃COO⁻Na⁺ + H₂O + CO₂

Moles in 0.3576g of NaHCO₃ (Molar mass: 84g/mol) are:

0.3576g of NaHCO₃ × (1mol / 84g) = <em>0.0042571 moles of NaHCO₃. </em>As 1 mole of NaHCO₃ reacts with 1 mole of CH₃COOH: <em>0.0042571 moles of CH₃COOH</em>

<em />

As molar mass of CH₃COOH is 60g/mol:

0.0042571 moles of CH₃COOH × (60g / 1mol) = <em>0.2554g of acetic acid is the minimum mass you would need for a full reaction</em>.

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In May 2016, William Trubridge broke the world record in free diving (diving underwater without the use of supplemental oxygen)

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Answer:

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

Explanation:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 3.6 L

V₂ = ?

P₁ = 1.0 atm

P₂ = 13.3 atm (From correct source)

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${1.0\ atm}\times {3.6\ L}={13.3\ atm}\times {V_2}$

${V_2}=\frac{{1.0}\times {3.6}}{13.3}\ L$

${V_2}=0.27\ L$

The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.

7 0

2 years ago

Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis

almond37 [142]

Answer: a) $BaCl_2$

b) $BiO_3H_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles$

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = $\frac{0.48}{0.48}=1$

For O = $\frac{0.96}{0.48}=2$

The ratio of Ba: Cl= 1:2

Hence the empirical formula is $BaCl_2$

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= $\frac{0.38}{0.38}=1$

For O = $\frac{1.15}{0.38}=3$

For H= $\frac{1.16}{0.38}=3$

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is $BiO_3H_3$

6 0

2 years ago

What is the mass of 1.82 moles of lithium carbonate?

Svetlanka [38]

You can view more details on each measurement unit: molecular weight of Lithium Carbonate or grams The molecular formula for Lithium Carbonate is Li2CO3. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles Lithium Carbonate, or 73.8909 grams.

5 0

2 years ago

Read 2 more answers

Given: CaC2 + N2 → CaCN2 + C In this chemical reaction, how many grams of N2 must be consumed to produce 265 grams of CaCN2? Exp

weeeeeb [17]

Answer : The grams of $N_2$ consumed is, 89.6 grams.

Solution : Given,

Mass of $CaCN_2$ = 265 g

Molar mass of $CaCN_2$ = 80 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate the moles of $CaCN_2$ .

$\text{Moles of }CaCN_2=\frac{\text{Mass of }CaCN_2}{\text{Molar mass of }CaCN_2}=\frac{265g}{80g/mole}=3.2moles$

The given balanced reaction is,

$CaC_2+N_2\rightarrow CaCN_2+C$

from the reaction, we conclude that

As, 1 mole of $CaCN_2$ produces from 1 mole of $N_2$

So, 3.2 moles of $CaCN_2$ produces from 3.2 moles of $N_2$

Now we have to calculate the mass of $N_2$

$\text{Mass of }N_2=\text{Moles of }N_2\times \text{Molar mass of }N_2$

$\text{Mass of }N_2=(3.2moles)\times (28g/mole)=89.6g$

Therefore, the grams of $N_2$ consumed is, 89.6 grams.

5 0

2 years ago

Read 2 more answers

When a 1.00 L sample of water from the surface of the Dead Sea (which is more than 400 meters below sea level and much saltier t

Harlamova29_29 [7]

Answer: Molarity of $MgCl_2$ in the original sample was 1.96M

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{\text{no of moles}}{\text{Volume in L}}$

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{186g}{95g/mol}=1.96$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1.96}{1.00L}$

$Molarity=1.96mol/L$

Thus molarity of $MgCl_2$ in the original sample was 1.96M

4 0

2 years ago