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JulsSmile [24]
3 years ago
8

"how a trait appears, or is physically expressed?

Chemistry
1 answer:
grin007 [14]3 years ago
6 0
I think the answer is genotype.
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A 48.3 mL sample of gas in a cylinder is warmed from 22 C, 87 C to 0.0 C what is the final temperature
snow_lady [41]

The answer is:

the volume stays the same. it is the pressure that increases

7 0
3 years ago
How many moles of na2so4 are present in 284.078 grams of na2so4?<br> a. 2<br> b. 3<br> c. 5<br> d. 6
schepotkina [342]
There are 2 moles of NA2SO4 present in 284.078 grams of NA2SO4. Why? Well, we already know here that there are 284.078 grams of NA2SO4. To find how many moles are present in this compound, we would have to divide these grams by the molar mass of the compound. When we go to the periodic table of elements, we will find that there are two moles sodium (which represents 45.98 grams), one mole sulfur (which represents 32.06 grams), and four moles oxygen (which represents 64 grams). When combining these, we would get one mole of sodium sulfate which is representative of 142.04 grams total. That is our molar mass. Now that we have the molar mass, we can go back and divide. 284.078 divided by 142.04 is equal to about 1.99999, and that rounds up to 2 moles. Hence, there are 2 moles of NA2SO4 present in 284.078 grams of NA2SO4.

Your final answer: A is your answer.
3 0
2 years ago
Where is most of the mass found in an atom
IgorLugansk [536]
massive livand that sarah or someone is how u do it
7 0
3 years ago
Read 2 more answers
Can u help me with this
Anastasy [175]
The surface area would be 25cm, while the vloume would be
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6 0
3 years ago
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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
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