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astraxan [27]
2 years ago
11

C D

Chemistry
1 answer:
Helga [31]2 years ago
5 0
I think it’s tissue I just did it
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Which of the following is an example of how a scientist might use a model?
anzhelika [568]

D Microscopes

because they use them for DNA

a,b,and c, all you an item that can be used and not tested with.

4 0
3 years ago
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How many atoms of carbon are present in 1.5 mole of carbon.
umka2103 [35]

Answer:

there is one carbon atom and there are two oxygen atoms. So, in total there are three atoms.

Explanation:

5 0
3 years ago
A weather balloon has a volume of 105L at 0.97 atm when the temperature is 318K. What is the volume
chubhunter [2.5K]

Answer:

89 L

Explanation:

Step 1: Given data

  • Initial pressure (P₁): 0.97 atm
  • Initial volume (V₁): 105 L
  • Initial temperature (T₁): 318 K
  • Final pressure (P₂): 1.05 atm
  • Final volume (V₂): ?
  • Final temperature (T₂): 293 K

Step 2: Calculate the final volume of the weather balloon

If we assume that the gas inside the balloon behaves as an ideal gas, we can calculate the final volume of the gas using the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂ / T₁ × P₂

V₂ = 0.97 atm × 105 L × 293 K / 318 K × 1.05 atm = 89 L

3 0
2 years ago
What is the salinity seawater?
BigorU [14]
<span>Seawater is water from a sea or ocean. On average, seawater in the world's oceans has a salinity of approximately 3.5% or 35 parts per thousand. This means that for every 1 liter (1000 mL) of seawater there are 35 grams of salts (mostly, but not entirely, sodium chloride) dissolved in it.
I used google
</span>
7 0
3 years ago
A 13.97 g sample of nabr contains 22.34% na by mass. Considering the law of constant composition (definite proportions) how many
Grace [21]

As per the law of constant composition, a given sample will always contain the same number of elements that combine in the same mass proportion.

Therefore if a sample of 13.97 g of NaBr contains 22.39 % of Na by mass then,  a sample of 5.75 g of NaBr would also contain 22.39% Na by mass

Hence:

Mass of Na = 5.75 g * 22.39/100 = 1.287 g

5.75 g of NaBr would contain 1.29 g of Na

3 0
3 years ago
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