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nevsk [136]
3 years ago
11

A prostar has a high enough ________ to become a star

Chemistry
2 answers:
avanturin [10]3 years ago
6 0
The answer to your question is B) mass
   A prostar has a high enough MASS to become a star.
                                            Hope this helps, love to help:).
BigorU [14]3 years ago
5 0
The correct answer is A. Mass. Hope I helped! :)
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Given the chemical formulas of the following compounds, name each compound and state the rules you used to determine each name.
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5 0
3 years ago
Which of these is not a structural adaptation? A) Horns B) Thick white fur C) Seasonal migration D) A long beak
Snezhnost [94]

Answer:

c) Seasonal migration

Explanation:

Because A, B and C are all examples of structural adaptations. And seasonal migration is not something that is in an animals structure but is instinct.

Have a great day!

6 0
3 years ago
How much 0.85 M HCl solution can be made by diluting 250 mL of 10 M HCl?
7nadin3 [17]

<span>Answer is:  2940 mL of the HCL solution.</span>

c₁(HCl) = 10.0 M.

V₂(AgNO₃<span>) = ?.
c</span>₂(AgNO₃<span>) = 0.85 M.
V</span>₁(AgNO₃<span>) = 250 mL </span>÷ 1000 mL/L = 0.25 L.

<span> c</span>₁<span> - original concentration of the solution, before it gets diluted.
c</span>₂<span> - final concentration of the solution, after dilution.
V</span>₁<span> - volume to be diluted.
V</span>₂<span> - final volume after dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
V</span>₂(HCl) = c₁ · V₁ ÷ c₂.

<span> V</span>₂(HCl) = 10 M · 0.25 L ÷ 0.85 M.

<span> V</span>₂(HCl) = 2.94 L · 1000 mL = 2940 mL.

7 0
3 years ago
Read 2 more answers
How much carbon dioxide will 64 grams of ch4 is burned in oxygen?
timofeeve [1]
2.4368 MILOGRAMS  is burned
6 0
3 years ago
What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? a: 0.24 b: 13.1 c: 0.076 d: 6.55 e: 14.3
Len [333]

Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

When an acid is neutralized, the amount of base added must equal the amount of acid initially present. This base quantity is said to be the equivalent quantity. In other words, at the equivalence point the stoichiometry of the reaction is exactly fulfilled (there are no limiting or excess reagents), therefore the numbers of moles of both will be in stoichiometric relationship. So:

V acid *M acid = V base *M base

where V represents the volume of solution and M the molar concentration of said solution.

In this case:

  • V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
  • M acid= 0.129 M
  • V base= ?
  • M base= 0.135 M

Replacing:

0.0137 L* 0.129 M= V base* 0.135 M

Solving:

V base=\frac{0.0137 L*0.129 M}{0.135 M}

V base=0.0131 L = 13.1 mL

<u><em> The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).</em></u>

4 0
3 years ago
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