Guanine is one example of nitrogeous base that is a...?

7.5 moles of nitrogen gas (N2) is formed in the following reaction. How many grams

AfilCa [17]

Answer:

Mass = 255 g

Explanation:

Given data:

Number of moles of nitrogen = 7.5 mol

Mass of ammonia formed = ?

Solution:

Chemical equation:

3H₂ + N₂ → 2NH₃

Now we will compare the moles of nitrogen and ammonia.

N₂ : NH₃

1 : 2

7.5 : 2/1×7.5 = 15

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 15 mol × 17 g/mol

Mass = 255 g

6 0

3 years ago

Single stranded___ molecules translate he code for making protein

Radda [10]

Answer:

RNA

Explanation:

Central dogma is the process of the formation of protein from the sequence of nucleotide on DNA. It consists of two further processes Transcription and translation.

During Transcription, the information from the DNA sequence is transcripted in a sequence of RNA, that is single stranded and contains Uracil instead of Thymine. During Translation, the information from the RNA sequence is translated into a sequence of amino-acid, and a protein.

Hope it helps!

6 0

3 years ago

How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?

Gnoma [55]

Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

4 0

3 years ago

Ne, c4h10, is a component of natural gas that is used as fuel for cigarette lighters. the balanced equation of the complete comb

Simora [160]

<span>3.68 liters First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements: Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286 Looking at the balanced equation for the reaction which is 2 C4H10(g)+13 O2(g)â†’8 CO2(g)+10 H2O(l) It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have: 0.037851286 mol * 4 = 0.151405143 mol The ideal gas law is PV = nRT where P = Pressure V = Volume n = number of moles R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) ) T = absolute temperature (23C + 273.15K = 296.15K) So let's solve the formula for V and the calculate using known values: PV = nRT V = nRT/P V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm) V = (3.679338871 L*atm)/(1 atm) V = 3.679338871 L So the volume of CO2 produced will occupy 3.68 liters.</span>

4 0

3 years ago

Need help with this question using dimensional analysis

m_a_m_a [10]

The speed limit would be 70.2 mi/h

3 0

3 years ago

Guanine is one example of nitrogeous base that is a...?​

Guanine is one example of nitrogeous base that is a...?