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larisa86 [58]
3 years ago
15

if you weigh the lauric acid to the nearest 0.1g and you weigh the unknown solid to the nearest 0.0001g, can you calculate a mol

ar mass of your unknown solid to three significant figures? Show clearly your reasoning.
Chemistry
1 answer:
timama [110]3 years ago
3 0

Answer & Explanation:

The limiting significant figure in a calculation is our smallest figure. 1 g is one figure by using the significant figure rules. In the same manner we don't count zeros unless there is a non-zero number in front of it so .0001 g is still 1 figure.

So no we only know the unknown to one significant figure.

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Plzzzzzz helpppp!!!!!!
andreyandreev [35.5K]

Answer:

10.5 miles

Explanation:

7x1.5=10.5

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3 years ago
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A sample of CO2 gas at 100 degrees Celsius has a volume of 250 mL at 760 mm Hg. How many moles of CO2 are present
Fed [463]

There are 8.16 × 10-³ moles of CO2 gas at 100°C with a volume of 250 mL at 760 mm Hg.

HOW TO CALCULATE NUMBER OF MOLES:

The number of moles of a sample of gas can be calculated using the following formula:

PV = nRT

Where;

  • P = pressure of gas (atm)
  • V = volume (L)
  • n = number of moles (mol)
  • R = gas law constant (0.0821 Latm/molK)
  • T = temperature (K)

According to this question;

  • P = 760mmHg = 1 atm
  • T = 100°C = 100 + 273 = 373K
  • V = 250mL = 0.250L
  • n = ?

1 × 0.250 = n × 0.0821 × 373

0.250 = 30.62n

n = 0.250 ÷ 30.62

n = 8.16 × 10-³mol

Therefore, there are 8.16 × 10-³ moles of CO2 gas at 100°C with a volume of 250 mL at 760 mm Hg.

Learn more about number of moles at: brainly.com/question/4147359

6 0
3 years ago
what would be the ph of an aqueous solution of sulphuric acid which is 5×10^_5 mol l^_1 in concentration​
guajiro [1.7K]

Answer:

the ph of an aqueous solution of sulphuric acid which is 5*10^5 mol in concentration is basic in nature

6 0
3 years ago
Write the example of solid liquid and gas separately ​
zysi [14]

Answer:

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7 0
2 years ago
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The mass percent of Cl⁻ in a seawater sample is determined by titrating 25.00 mL of seawater with AgNO₃ solution, causing a prec
Serga [27]

Answer:

2.5 %

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For AgNO_3 :

Molarity = 0.2850 M

Volume = 63.30 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 63.30 × 10⁻³ L

Thus, moles of AgNO_3 :

Moles=0.2850 \times {63.30\times 10^{-3}}\ moles

Moles of AgNO_3  = 0.0180405 moles

Moles of AgNO_3  = Moles of Cl^-

Thus, Moles of Cl^- = 0.0180405 moles

Molar mass of Cl^- = 35.453 g/mol

Mass = Moles * Molar mass = 0.0180405 moles * 35.453 g/mol = 0.6396 g

Volume of sea water = 25.00 mL

Density = 1.024 g/mL

Density = Mass / Volume

Mass = Density * Volume = 1.024 g/mL * 25.00 mL = 25.6 g

Mass\ \%=\frac{Mass_{Chloride ion}}{Total\ mass}\times 100

Mass\ \%=\frac{0.6396}{25.6}\times 100

<u>Mass percent of Cl⁻ = 2.5 %</u>

4 0
3 years ago
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