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Alexandra [31]
2 years ago
7

Using the Cell Size and Scale Interactive, tell me which of the following is correct. They are in order of size, left to right,

from smallest to largest:
tRNA > influenza virus > E. coli bacterium
Hepatitis virus > hemoglobin > phospholipid
Ribosome > amoeba proteus > carbon atom
Chemistry
1 answer:
Mkey [24]2 years ago
3 0

Using the cell size and scale Interactive, the correct order of size, left to right, from smallest to largest is tRNA > influenza virus > E. coli bacterium Hepatitis virus > hemoglobin > phospholipid. The correct option is a.

<h3>What is t-RNA?</h3>

t-RNA is transfer RNA. It helps in transferring information from RNA to DNA or DNA to RNA.

t-RNA is present in viruses, so it will be smaller than the virus because they are present inside the virus. Phospholipids are chains of lipids. They are macromolecules.

Therefore, the correct option is a. tRNA > influenza virus > E. coli bacterium Hepatitis virus > hemoglobin > phospholipid.

To learn more about Cell Size and Scale Interactive, refer to the link:

brainly.com/question/917170

#SPJ1

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A sample of carbon dioxide is contained in a 250.0 mL flask at 0.930 atm and 15.4 °C. How many molecules of gas are in
Yanka [14]

Answer:

We are given:

Volume (V) = 0.25 L

Pressure (P) = 0.93 atm

Temperature (T) = 15.4°C  OR   288.4 K

<u>Solving for the number of moles of CO₂:</u>

From the ideal gas equation:

PV = nRT

replacing the variables

0.93 * 0.25 = n (0.082)(288.4)

n = 0.00983 moles

<u>Number of molecules:</u>

Number of moles=  0.00983

number of molecules in 1 mole = 6.022 * 10²³

Number of molecules in 0.00983 moles = 0.00983 * 6.022 * 10²³

Number of molecules = 5.91 * 10²¹

8 0
3 years ago
The equilibrium constant has been estimated to be 0.12 at 25 °C. If you had originally placed 0.069 mol of cyclohexane in a 2.8
scZoUnD [109]

Answer: Concentrations of cyclohexane and methylcyclopentane at equilibrium are 0.0223 M and 0.0027 M respectively

Explanation:

Moles of cyclohexane = 0.069 mole

Volume of solution = 2.8 L

Initial concentration of cyclohexane =\frac{moles}{Volume}=\frac{0.069}{2.8}=0.025M

The given balanced equilibrium reaction is,

                            cyclohexane  ⇔  methylcyclopentane

Initial conc.                 0.025 M           0

At eqm. conc.       (0.025-x)M       (x) M

The expression for equilibrium constant for this reaction will be,

K= methylcyclopentane / cyclohexane

Now put all the given values in this expression, we get :

0.12=\frac{(x)}{(0.025-x)}

By solving the term 'x', we get :

x =  0.0027

Concentration of cyclohexane at equilibrium = (0.025-x ) M = (0.025-0.0027) M = 0.0223 M

Concentration of methylcyclopentane at equilibrium = (x ) M = (0.0027) M

4 0
4 years ago
Maya is playing the guitar. She drums the strings and the guitar procedures noise. maya is transforming____energy into _____ener
posledela
The strumming is mechanical energy
the notes heard is sound energy
so use D
6 0
3 years ago
List two detection (i.e. visualisation) techniques commonly used to visualise<br> compounds in TLC.
olya-2409 [2.1K]

Answer:

The most common non-destructive visualization method for TLC plates is ultraviolet (UV) light. A UV lamp can be used to shine either short-waved (254nm) or long-waved (365nm) ultraviolet light on a TLC plate with the touch of a button

Explanation:

hope this helps

7 0
3 years ago
Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175, 104, 164, 193, 13
Mila [183]

Answer:

Following are the solution to these question:

Explanation:

Calculating the mean:

\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}\\\\

  =\frac{1571}{10}\\\\=157.1

Calculating the standardn:

\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\\\\

Please find the correct equation in the attached file.

=28.195

For point a:

=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585\\

For point b:

=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M

For point c:

= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\  M

It is calculated by using the slope value that is 1.75 \times 10^9 M^{-1}. The slope value 1.75 \times 10^9 M^{-1}is ambiguous.

7 0
2 years ago
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