Answer:
metals, nonmetals and metalloids
Explanation:
The domain level divides
organisms based on the
differences between
eukaryotes and
prokaryotes. Eukaryotes
have a nucleus in their
cells. Which BEST describes
prokaryotes?
A. Prokaryotes are nonliving things.
B. Prokaryotes do NOT have a nucleus.
C. Prokaryotes include all plant life.
D. Prokaryotes cannot reproduce.
Answer:
B. Prokaryotes do NOT have a nucleus.
How many minutes are in 180,0 days?
Answer:
259,200 minutes
Explanation:
What causes warm air to rise?
a. The fact that it's less dense than cold air.
b. The fact that it weighs more than cold air.
c. The fact that it has higher pressure than cold air.
d. The fact that it's more dense than cold air.
1.25 x 10^12 molecules H2SO4 to moles
2.08 × 10⁻¹² mol H₂SO₄
General Formulas and Concepts:Math
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightChemistry
Atomic Structure
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.Stoichiometry
Using Dimensional AnalysisExplanation:Step 1: Define
1.25 × 10¹² molecules H₂SO₄
Step 2: Identify Conversions
Avogadro's Number
Step 3: Convert
Set up: [tex]\displaystyle 1.25 \cdot 10^{12} \ molecules \ H_2SO_4(\frac{1 \ mol \ H_2SO_4}{6.022 \cdot 10^{23} \ molecules \ H_2SO_4})[/tex]Multiply/Divide: [tex]\displaystyle 2.07572 \cdot 10^{-12} \ moles \ H_2SO_4[/tex]Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
2.07572 × 10⁻¹² mol H₂SO₄ ≈ 2.08 × 10⁻¹² mol H₂SO₄
the substance that can not be further decomposed by ordinary chemical means is
A.water
B.gold
C.sugar
D.air
If a section of DNA has 15% thymine, how much adenine will it have?
1. 15%
2. 30%
3. 60%
4.85%
Answer:
15% adenine
Explanation:
Thymine and adenine are a nitrogenous base pair so they would equal the same.
Please help! I really need help on this!
Answer:
I wanna say c
Explanation:
It makes the most sense, because non-metals don't produce an electrical current, and therefore aren't conductors.