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sergey [27]
4 years ago
14

The general name for hydrocarbons with at least one triple covalent bond is

Chemistry
2 answers:
melomori [17]4 years ago
6 0
Alkynes are hydrocarbons with at least one triple covalent bond.
timurjin [86]4 years ago
3 0

Answer:

Alkynes

Explanation:

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Enter a balanced equation for the dissolution of caco3.
Tatiana [17]

Answer:

CaCO3(aq) → Ca2+(aq ) + CO3 2-(aq)

Explanation:

  • Dissolution reactions are reactions that occur when a solute either in gaseous, liquid, or solid form dissolves in a solvent such as water to form a solution.
  • In this case we are given Calcium carbonate (CaCO3) which undergoes dissolution according to the equation;
  • CaCO3(aq) → Ca2+(aq ) + CO3 2-(aq)
  • Then<em><u> the bicarbonate ion combines with two protons from water to form a weak acid H2CO3. The weak acid is then broken down to form CO2 and H2O since its unstable.</u></em>
5 0
3 years ago
A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.
dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution =  9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

3 0
3 years ago
Is vineasel edible??????????????????????/
nevsk [136]

Answer:

no it's not edible

5 0
3 years ago
Read 2 more answers
An independent variable is....
Irina18 [472]

Answer:

B - What we change

Explanation:

Dependent Variable - What we measure

Control Variable - what stays the same

Conclusion - what we conclude

<em>Hope</em><em> </em><em>this</em><em> </em><em>can</em><em> </em><em>Help</em><em>!</em>

<em>:</em><em>D</em>

5 0
3 years ago
A piece of metal with a mass of 611 g is placed into a graduated cylinder that contains 25.1 ml of water, raising the water leve
jeka94

Mass of metal piece is 611 g and volume of graduated cylinder is 25.1 mL. When metal piece is placed in the graduated cylinder water level increases to 56.7 mL. The increase in volume is due to volume of metal piece that gets added to the volume of water.

Thus, volume of metal piece can be calculated by subtracting initial volume from the final one.

V_{metal}=V_{final}-V_{initial}=(56.7-25.1)mL=31.6 mL

Thus, volume of metal piece will be 31.6 mL. The mass of metal piece is given 611 g, density of metal can be calculated as follows:

d=\frac{m}{V}=\frac{611 g}{31.6 mL}=19.33 g/mL

Therefore, density of metal is 19.33 g/mL.

3 0
3 years ago
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