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Nucleotides have a phosphate group attached at the ___ carbon atom of the sugar.

Tanzania [10]

Answer:

Nucleotides have a phosphate group attached at the 5' carbon atom of the sugar.

Explanation:

Phosphate group is attached to the sugar molecule in the place of -OH.

6 0

3 years ago

What is a time machine? who discovered the time machine

o-na [289]

A time machine is a device which can transport an object or person back in time or in to the future. 

Ali Razeghi is the person who invented the time machine.

6 0

3 years ago

How many grams are in 3.00 x 10^24 molecules of Zn(C2H3O2)2?

Burka [1]

Answer:

Mass = 917.4 g

Explanation:

Given data:

Number of molecules = 3.00 ×10²⁴ molecules

Mass in gram = ?

Solution:

1 mole contain 6.022 × 10²³ molecules

3.00 ×10²⁴ molecules × 1 mol / 6.022 × 10²³ molecules

0.5×10¹ mol

5 mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 5 mol × 183.48 g/mol

Mass = 917.4 g

7 0

3 years ago

A laboratory analysis of an unknown sample yields 40.5% carbon, 4.5% hydrogen, and 55.0% oxygen. if the sample has a molar mass

den301095 [7]

Answer is: molecular formula is C₆H₈O₆.
n(C) = m(C) ÷ M(C).
n(C) = 40,5 g ÷ 12 g/mol.
n(C) = 3,375 mol.
n(H) = m(H) ÷ M(H).
n(H) = 4,5 g ÷ 1 g/mol.
n(H) = 4,5 mol.
n(O) = m(O) ÷ M(O).
n(O) = 55 g ÷ 16 g/mol.
n(O) = 3,4 mol.
n(C) : n(H) : n(O) = 3,375 mol : 4,5 mol : 3,4 mol / :3,375.
n(C) : n(H) : n(O) = 1 : 1,33 : 1.
n(C) : n(H) : n(O) = 3 : 4 : 3.
M(C₃H₄O₃) = 88 g/mol · 2 = 176 g/mol.

6 0

3 years ago

What change in volume results if 60 mL of a gas is cooled from 33 C to 5 C?

Eva8 [605]

Answer:

Change in volume on changing temperature from 33 $^{\circ}C$ to 5 $^{\circ}C$ is 5.49 mL

Explanation:

Initial volume of gas = V = 60 mL

Assuming final volume of gas to be V' mL

Initial temperature = T = 33 $^{\circ}C$ = 306 K

Final temperature = T' = 5 $^{\circ}C$ = 278 K

The relationship between volume and temperature of gas at constant pressure is shown below

$\displaystyle \frac{V}{V'}=\displaystyle \frac{T}{T'} \\\displaystyle \frac{60\textrm{ mL}}{V'} = \displaystyle \frac{306\textrm{ K}}{T} \\V' = 54.51 \textrm{ mL} \\\textrm{Change in volume} = \left ( V-V' \right ) \\\textrm{Change in volume} = \left ( 60-54.51 \right )\textrm{ mL} \\\textrm{Change in volume} = 5.49 \textrm{ mL}$

Change in volume on changing temperature = 5.49 mL

6 0

3 years ago