Answer:
phenotype,phenotype,genotype,genotype,
genotype
Explanation:
phenotype is physical appearance and genotype is just like
yy Tt
Answer:
A. is alloyed with carbon and chromium to make stainless steel.
Explanation:
Steel is an alloy formed mainly of iron and carbon but some other metals like chromium are also added in little amounts.
Is steel, the percentage of iron and carbon together is about 90% and the rest metals fall in the 10% part.
Although the cost of steel is low, it has a very high tensile strength and that's why it is used in tools, ships, buildings, trains and in various types of infrastructures.
There are four states of matter, solid, liquid, gas and plasma. Their formation is as when solid is heated it converts into liquid, liquid on heating converts into gases and gases on heating converts into plasma.
Plasma:
Plasma is the fourth state of matter. It is the highest energy state of matter.
Composition:
Plasma is made up of negatively charged and positively charged particles.
Result:
The answer to your question is Plasma.
<h3>
Answer:</h3>
0.127 mol Au
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 25.0 g Au
[Solve] moles Au
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.126923 mol Au ≈ 0.127 mol Au
Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
