Answer:
38.16 mL
Solution:
Data Given;
Density = 1.31 g/mL
Mass = 50 g
Volume = ?
Formula Used;
Density = Mass ÷ Volume
Solving for Volume,
Volume = Mass ÷ Density
Putting values,
Volume = 50 g ÷ 1.31 g.mL⁻¹
Volume = 38.16 mL
Answer:
The SAE curriculum includes practical farming tasks conducted outside the scheduled classroom and laboratory period by students. SAEs offer a method for students in agricultural education to gain real-world work opportunities that they are most interested in in the field of agriculture. Supervised agricultural experience is an essential component of agricultural education, and all Agriculture, Food and Natural Resources (AFNR) courses are a necessary component.
Explanation: Hope it helps
Answer:
I think a red flower with one part colored yellow
Explanation:
We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)
The specific heat of the metal is 0.44 J/ (°C × g)
<h3>
Answer:</h3>
2 L Ne
<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>
- Using Dimensional Analysis
- STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
0.07 mol Ne (g)
<u>Step 2: Identify Conversions</u>
STP - 22.4 L per mole
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
1.568 L Ne ≈ 2 L Ne
