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Bond [772]
3 years ago
15

Pre-lab Assignment

Chemistry
1 answer:
alina1380 [7]3 years ago
3 0

Answer:

1. d. Person A rounded their measurement

2) a. 105.

b. 76.4

c. 0.04

d. 1.988

Explanation:

1. Measuring an object is done with the least precise one. Person A recorded 1.5cm while Person B recorded 1.50cm of the same object. This measurement differs in the sense that Person A rounded up their measurement to 2 significant figures (1.5) while Person B recorded their answer as three significant figures (1.50).

2. During multiplication and division, the result is written as the measurement in the lowest significant figure.

a. 7.00 x 15.00 = 105. Calculator showed 105.00, which is 5 s.f (two zeros after decimal point are significant). The least significant figure is 3, hence 105.00 is rounded up to 105.

b. 40.2.1.901 = 76.4; The calculator answer of this operation is 76.420200 but since the lowest significant figure is 3. We round up our answer to 3 s.f i.e. 76.4

c. (0.003)(13.2) = 0.04. For this operation, calculator answer is 0.0396, however, the lowest significant figure in the question is 1 i.e. 0.003. Note that the two zeros before 3 are insignificant. Hence, our answer will be rounded up to 1 s.f i.e. 0.04

d. 1.590+ 0.3975 = 1.988. the calculator answer to this operation is 1.9875. However, the least significant figure in the question is 4 i.e. 1.590. hence, the answer will be rounded up to 4 s.f which is 1.988.

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DedPeter [7]

Answer:

Moles to Grams caco3

1 mole is equal to 1 moles CaCO3, or 100.0869 grams.

6 0
3 years ago
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A sample of gas in a 14.6 L flexible container is at 25.0oC and 1.00atm. What is the volume of the sample when heated to 220.0oC
inn [45]

Answer: 24.1 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=14.6L\\T_1=25.0^oC=(25+273)K=298K\\V_2=?\\T_2=220.0^0C=(220+273)K=493K

Putting values in above equation, we get:

\frac{14.6}{298K}=\frac{V_2}{493}\\\\V_2=24.1L

Thus the volume of the sample when heated to 220.0oC and the pressure is constant is 24.1 L

7 0
3 years ago
How many moles of ethanol are produced starting with 500.g glucose?
Monica [59]
<h3>Answer:</h3>

5.55 mol C₂H₅OH

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Tables
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                 \displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})
  2. [DA} Multiply/Divide [Cancel out units]:                                                         \displaystyle 5.55001 \ mol \ C_2H_5OH

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

8 0
3 years ago
Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)
cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

3 0
3 years ago
When molecules are joined I an ordered structure​
olga2289 [7]

Answer:

ice

Explanation:

5 0
3 years ago
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