<h3>
Answer:</h3>
2.0 mol C₆H₁₂O₆
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
1.2 × 10²⁴ molecules C₆H₁₂O₆ (glucose)
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
- Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
1.99269 mol C₆H₁₂O₆ ≈ 2.0 mol C₆H₁₂O₆
Answer:
Acceleration is zero.
Explanation:
The slope of a position time graph gives the velocity of the body.
If the slope is constant means the velocity is constant.
Now, acceleration is the measure of the change in velocity of a body over a given time interval.
So, the acceleration of a body is directly proportional to the change in velocity of the body.
If there is no change in velocity, this means that the acceleration of the body is zero.
Here, the slope is a constant implying that the velocity is a constant. So, there is no change in velocity. This implies that the acceleration is zero for the body in the given time interval.
Thus, if a position time graph has a constant slope, one can infer that the acceleration is zero.
Answer:
cancer between 6figures yard is day awl
Answer:
There was an improvement in accuracy. There was no change in precision.
Explanation:
<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).
The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.
Answer:
- 130.64°C.
Explanation:
- We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 634.0 L, T₁ = 21.0°C + 273 = 294.0 K.
V₂ = 307.0 L, T₂ = ??? K.
<em>∴ T₂ = V₂T₁/V₁ </em>= (307.0 L)(294.0 K)/(634.0 L) = <em>142.36 K.</em>
<em>∴ T₂(°C) = 142.36 K - 273 = - 130.64°C.</em>
