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Alborosie
3 years ago
8

What are some example of nuclear acids

Chemistry
1 answer:
vladimir1956 [14]3 years ago
3 0
First of all, it is a nucleic acid, not nuclear acid.

Examples are DNA and RNA( Deoxyribonucleic Acid and Ribonucleic Acid)
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A glucose solution that is prepared for a patient should have a concentration of 180 g/L. A nurse has 18 g of glucose. How many
Anuta_ua [19.1K]
You start by using proportions to find the number of liters of solution:

180 g of glucose / 1 liter of solution = 18 g of glucose / x liter of solution


=> x = 18 g of glucose * 1 liter of solution / 180 g of glucose = 0.1 liter of solution.


If you assume that the 18 grams of glucose does not apport volume to the solution but that the volume of the solution is the same volumen of water added (which is the best assumption you can do given that you do not know the how much the 18 g of glucose affect the volume of the solution) then you should add 0.1 liter of water.

Answer: 0.1 liter of water.

 
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3 years ago
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Greg has 36 DVDs in his collection.
sp2606 [1]

Answer:

14

Explanation:

4 0
4 years ago
Why are the Flemish called Flemish
Dmitriy789 [7]

Explanation:

The flemish are an ethnic group which belong to the northern part of Belgium called Flanders. They also speak a language called flemish. The major population of Belgium are the flemish. The language they speak is a  Low Franconian dialect cluster of the Dutch language.

There was a country called the flanders located in the lowlands of Europe. Any person from this country was considered as flemish previously.

4 0
3 years ago
True of False: Only conclusions that support the hypothesis should be reported in scientific research.
RSB [31]
False, If something doesn't support your hypothesis maybe it was wrong in the first place and you should make it known in your research if you come up with a different conclusion.
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A sample of xenon occupies a volume of 736 mL at 2.02 atm and 1 °C. If the volume is changed to 416 mL and the temperature is ch
kozerog [31]

Answer:

\large \boxed{\text{4.63 atm}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}

Data:

p₁ = 2.02 atm; V₁ = 736 mL; n₁ = n₁; T₁ =    1 °C

p₂ = ?;             V₂ = 416 mL; n₂ = n₁; T₂ =  82 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (   1 + 273.15) K = 274.15 K

T₂ = (82 + 273.15) K = 355.15 K

(b) Calculate the new pressure

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{\text{2.02 atm}\times \text{736 mL}}{n _{1}\times \text{274.15 K}} & = &\dfrac{p_{2}\times \text{416 mL}}{n _{1}\times \text{355.15 K}}\\\\\text{5.423 atm} & = &1.171{p_{2}}\\p_{2} & = & \dfrac{\text{5.423 atm}}{1.171}\\\\ & = & \textbf{4.63 atm} \\\end{array}\\\text{The  new pressure will be $\large \boxed{\textbf{4.63 atm}}$}}

6 0
3 years ago
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