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Naddik [55]
3 years ago
12

What is the number of electrons shared in the multiple carbon-carbon bond in one molecule of 1-pentyne?

Chemistry
1 answer:
jeka57 [31]3 years ago
8 0
Where the options are?
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Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
const2013 [10]

Answer:

The answer is 1.15m.

Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.

We can find the number of H2SO4 moles by using its molarity

C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288

Since water has a density of 1.00kgL, the mass of solvent is

m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg

Therefore, molality is

m=nmass.solvent=0.288moles0.250kg=1.15m

4 0
3 years ago
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As carbon dioxide enters systemic blood, it causes more oxygen to dissociate from hemoglobin, which in turn allows more CO2 to c
vagabundo [1.1K]

Answer:

The correct answer is "False".

Explanation:

It is false that as carbon dioxide enters systemic blood, it causes more oxygen to dissociate from hemoglobin. Once an atom of oxygen binds to hemoglobin, hemoglobin change its shape and makes easier than a second and a third atom of oxygen binds towards it. This change in conformation makes no possible that carbon dioxide can cause that oxygen dissociates from hemoglobin.

5 0
3 years ago
Which pair contains an incorrect formula.
Eva8 [605]
C) Sliver Carbonate AgCO3
3 0
3 years ago
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Based on these data, what is the average atomic mass of element b? 10.01 10.51 10.81 11.01
Lelechka [254]
The right answer is C. 10.81
5 0
3 years ago
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If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
kiruha [24]

Answer:

Sr(OH)₂ will be the limiting reagent.

Explanation:

First of all, you should know the following balanced chemical equation:

2 H₂CO₃ + 2 Sr(OH)₂ → 4 H₂O + Sr₂(CO₃)₂

The balanced equation is based on the Law of Conservation of Mass, which says that matter cannot be created or destroyed. Therefore, the number of each type of atom on each side of a chemical equation must be the same.      

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). By stoichiometry the following amounts in moles react:

  • strontium hydroxide: 2 moles
  • carbonic acid: 2 moles

Now, you know the following masses of the elements:

  • Sr: 87.62 g/mole
  • O: 16 g/mole
  • H: 1 g/mole

So the molar mass of strontium hydroxide is:

Sr(OH)₂= 87.62 g/mole + 2*(16 g/mole + 1 g/mole)= 121.62 g/mole

You apply the following rule of three, if 121.62 grams of hydroxide are present in 1 mole, 12.5 grams in how many moles are they?

moles of strontium hydroxide=\frac{12.5 grams*1 mole}{121.62 grams}

moles of hydroxide= 0.103 moles

On the other hand, you have 150 ml of 3.5 M carbonic acid. Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you can apply the following rule of three: if in 1 L there are 3.5 moles of carbonic acid, in 0.150 L (being 1 L = 1000 mL, 0.150 L = 150 mL) how many moles of acid are there?

molesofcarbonicacid=\frac{0.150 L*3.5 moles}{1 L}

moles of carbonic acid= 0.525 moles

Finally, to calculate the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 2 mole of strontium hydroxide reacts with , how much moles of carbonic acid will be needed if 0.103 moles of strontium hydroxide react?

molesofcarbonicacid=\frac{0.103 moles of strontium hydroxide*2 moles of carbonic acid}{2 moles of strontium hydroxide}

moles of carbonic acid= 0.103 moles

But 0.525 moles are available. Since more moles are available than you need to react with 0.103 moles of strontium hydroxide, <u><em>Sr(OH)₂ will be the limiting reagent.</em></u>

7 0
3 years ago
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