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2 years ago

Help me plss ty in advanced

Chemistry

1 answer:

Komok [63]2 years ago

3 0

Answer:

See below ~

Explanation:

The respective pairs are :

× Ribose ⇒ The sugar of RNA

× Double helix ⇒ Structures of DNA

× Watson & Crick ⇒ Proposed the double helix model

× Base pair ⇒ Adenine - Thymine pair

× Cytosine ⇒ The nucleotide pair of Guanine

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0.03÷0.35 what is it rounding at least 3 digits

strojnjashka [21]

0.0857
Rounding off to three digits
Since 7 is more than five you change it to 0 and add 1 to 5
A: 0.086

6 0

3 years ago

The diameter of a biscuit is approximately 51 millimeters (mm). An atom of bismuth (Bi) is approximately 320. picometers (pm) in

astraxan [27]

Answer:

1.5e+8 atoms of Bismuth.

Explanation:

We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):

$\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}$

For this, it is necessary to know the values in meters for any of these diameters:

$\\ 1m = 10^{3}mm = 1e+3mm$

$\\ 1m = 10^{12}pm = 1e+12pm$

Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.

<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>

1 atom of Bismuth = 320pm in diameter.

$\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m$

<h3>Diameter of a biscuit in meters</h3>

$\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m$

<h3>Resulting Ratio</h3>

How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}$

$\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}$

$\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8$

In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.

6 0

3 years ago

Lab report Motion: 3. What methods are you using to test this (or each) hypothesis?

nadya68 [22]

Answer: You can increase the weight, then test the speed, and make the weight normal and test the speed, and mark which one travels faster.

Explanation: This would test your hypothesis by comparing the speeds of the cars when more mass is added. Calculating the difference of the speed with more mass, and the speed with normal mass would give you your answer. A positive number would prove your hypothesis and a negative number would disprove it.

8 0

3 years ago

Weight of one mole of carbon = 12.01 g Weight of one mole of oxygen = 16.00 g The molecular weight (gram formula weight) for CO

murzikaleks [220]

Answer:

28.01g

Explanation:

Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.

The molecular weight of a compound can be gotten by adding the molar weights of the elements that constitutes the compound .

The molecular weight of the compound CO is therefore

equal to the sum of the weight of both elements.

That’s = 12.01g + 16.00g

= 28.01g

Therefore, the molecular weight of CO is 28.01g

4 0

4 years ago

10. A 2.36-gram sample of NaHCO3 was completely decomposed in an

Airida [17]

Answer:

0.79 g

Explanation:

Let's introduce a strategy needed to solve any similar problem like this:

Apply the mass conservation law (assuming that this reaction goes 100 % to completion): the total mass of the reactants should be equal to the total mass of the products.

Based on the mass conservation law, we need to identify the reactants first. Our only reactant is sodium bicarbonate, so the total mass of the reactants is:

$m_r=m_{NaHCO_3}=2.36 g$