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pantera1 [17]
3 years ago
13

A sample of nitrogen containing 3.0×10^23 molecules has the same number of molecules as a sample containing

Chemistry
1 answer:
Daniel [21]3 years ago
5 0

Answer:

D) 0.50 mole of Ne

Explanation:

Given data:

Number of molecules of nitrogen = 3.0×10²³ molecules

Which sample contain same number of molecules as nitrogen= ?

Solution:

A) 0.25 mole of O₂

1 mole = 6.022×10²³ molecules

0.25 mol × 6.022×10²³ molecules / 1 mol

1.51×10²³ molecules

B) 2.0 moles of He.

1 mole = 6.022×10²³ molecules

2.0 mol × 6.022×10²³ molecules / 1 mol

12.044×10²³ molecules

C) 1.0 moles of H₂

1 mole = 6.022×10²³ molecules

D) 0.50 mole of Ne

1 mole = 6.022×10²³ molecules

0.50 mol × 6.022×10²³ molecules / 1 mol

3.0×10²³ molecules

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Answer:

A pure trait is also known as a homozygous trait. Homozygous traits are either a combination of the same two dominant alleles or the same two recessive alleles. A hybrid trait is also known as a heterozygous trait, and is the pairing of a dominant and recessive allele.

Explanation:each possible combo has a term for it. When we have two capital or two lowercase letters in the GENOTYPE

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4 years ago
Which three of the following are steps of aerobic cellular respiration?
aleksley [76]
A.
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3 0
3 years ago
PLEASE HELP ME!!!
kondor19780726 [428]
<h3>Answer:</h3>

0.424 J/g °C

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Thermochemistry</u>

Specific Heat Formula: q = mcΔT

  • q is heat (in Joules)
  • m is mass (in grams)
  • c is specific heat (in J/g °C)
  • ΔT is change in temperature
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

<u>Step 2: Solve for Specific Heat</u>

  1. Substitute in variables [Specific Heat Formula]:                                             181 J = (38.8 g)c(11.0 °C)
  2. Multiply:                                                                                                             181 J = (426.8 g °C)c
  3. [Division Property of Equality] Isolate <em>c</em>:                                                         0.424086 J/g °C = c
  4. Rewrite:                                                                                                             c = 0.424086 J/g °C

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.424086 J/g °C ≈ 0.424 J/g °C

6 0
3 years ago
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Describe the three steps of solution formation. be sure to include the details regarding the dissolving of the substance
Elza [17]

The three steps involve;

Step 1: Separation/expansion of the solute particles

Step 2: Separation/expansion of the solvent particles

Step 3; Combining the solute and solvent particles


The first two steps are usually endothermic. Step 3, nonetheless, can be either exothermic or endothermic and is significant in determining whether the dissolving process will be endothermic or exothermic.


8 0
3 years ago
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