Answer:
Step-by-step explanation:
Remark
The formula for this is
Heat = m * c * Δt
Givens
Heat = unknown
m = 200 grams
c = 4.184 Joules / grams oC
Δt = the change in temperature = 35.7 - 22.3 = 13.4
Solution
Heat = 200 * 4.184 * 13.4
Heat = 11213 Joules
Heat = 11.2 Kj
It's hard to get the correct number of sig digs because the 200 is not qualified in any way.
Part 2.
This can really puzzle you until you know that heat given up = the heat taken on. The heat taken on was 11.2 Kj. The Butane lighter must have given that heat on.
Heat of Combustion = Heat Given Up / Number grams burned.
Heat of Combustion = 11.2 Kj / 0.23 grams
Heat of Combustion = 48.8 Kj / gram
Answer:
68
Step-by-step explanation:
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Answer:
(1, 3)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
<u>Algebra I</u>
- Coordinates (x, y)
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
y = 3
y = -3x + 6
<u>Step 2: Solve for </u><em><u>x</u></em>
- Substitute in <em>y</em>: 3 = -3x + 6
- [Subtraction Property of Equality] Subtract 6 on both sides: -3 = -3x
- [Division Property of Equality] Divide -3 on both sides: 1 = x
- Rewrite/Rearrange: x = 1
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define original equation: y = -3(1) + 6
- Multiply: y = -3 + 6
- Add: y = 3