To write quantities in ordinary notation, you need to notice the power of the exponent.
If the power is positive, then you move the decimal point to the right by the number the power in exponent tells you.
If the power is negative, then you move the decimal point to the left by the number the power in exponent tells you.
We have:
3 x 10^-4 : the power of exponent is negative, therefore, we will move the decimal point 4 places to the left.
3 x 10^-4 = 0.0003 km
3 x 10^4 : he power of exponent is positive, therefore, we will move the decimal point 4 places to the right.
3 x 10^4 = 30000 km
<h3>
Answer:</h3>
733 g CO₂
<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>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O
[Given] 5.55 mol C₃H₇OH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol C₃H₇OH → 6 CO₂
Molar Mass of C - 12.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol
<u>Step 3: Stoichiometry</u>
- Set up conversion:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
732.767 g CO₂ ≈ 733 g CO₂
Answer:
1) The Kelvin temperature cannot be negative
2) The Kelvin degree is written as K, not ºK
Explanation:
The temperature of an object can be written using different temperature scales.
The two most important scales are:
- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).
- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.
The expression to convert from Celsius degrees to Kelvin is:
Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:
1) The Kelvin temperature cannot be negative
2) The Kelvin degree is written as K, not ºK
Answer:
187.34 atm
Explanation:
From the question,
PV = nRT.................. Equation 1
Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.
make P the subject of the equation
P = nRT/V.............. Equation 2
n = mass(m)/molar mass(m')
n = m/m'............... Equation 3
Substitute equation 3 into equation 2
P = (m/m')RT/V............ Equation 4
Given: m = 46 g, T = 25°C = (25+273) = 298 K, V = 3.00 L
Constant: m' = 2 g/mol, R = 0.082 atmL/K.mol
Substitute these values into equation 4
P = (46/2)(0.082×298)/3
P = (23×0.082×298)/3
P = 187.34 atm
The best explanation would be that Gases were released during the Chemical reaction, causing a loss of Mass.