Answer:
butyne
Explanation:
alkane, alkene, and alkyne are all examples of hydrocarbons.
butyne = alkyne
Answer:
Gallium-72
Explanation:
The elements are identified by the number of protons of the atom, which is its atomic number.
In this case the number of protons 39 (atomic number 39) permit you to identify the element as gallium.
Now, to identify the isotope you tell the name of the element and add the mass number.
The mass number is the sum of the protons and the neutrons
In this case, the number of neutrons is the original 39 plus the 2 added suddenly, i.e. 39 + 2 = 41, so the mass number is 31 + 41 = 72
Therefore, the isotope is gallium - 72.
The concentration of the original calcium ions is 0.005 M
<h3>What is concentration?</h3>
The term concentration has to do with the amount of substance in solution. We know that the concentration can be measured in a lot of units such as mole/litre, grams per litre, percentage and so on.
As such we have the equation;
Ca^2+(aq) + (NH4)2CrO4(aq) --------> CaCrO4(s) + 2NH4^+(aq)
Number of moles of the precipitate = 346.7 * 10^-3 g/156 g/mol
= 0.0022 moles
Now;
1 mole of Ca^2+ produces 1 mole of CaCrO4 hence 0.0022 moles of CaCrO4 was produced by 0.0022 moles of CaCrO4.
Given that the volume of the solution is 0.440 L, the concentration of the solution is; 0.0022 moles/0.440 L
= 0.005 M
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The solubility KI is 50 g in 100 g of H₂O at 20 °C. if 110 grams of ki are added to 200 grams of H₂O <u>the </u><u>solution </u><u>will be </u><u>saturated</u><u>.</u>
<h3>What is solubility?</h3>
Solubility is a condition where the solute is fully dissolved in the solvent. When fully mixed with the solvent.
Given that 50 g of KI is added to 100 g of water at 20 °C it means 100 g of water can dissolve a maximum of 50 g of KCl.
1 g of water will dissolve an quantity of 0.5 g of KCl.
To assay for 200 g of water: 200 g of water can disintegrate a maximum of (0.5) x 200 g of KCl.
The maximum amount of KCl that will dissolve is 100 g
Actualised amount dissolved = 110 g
when Amount dissolved > Maximum solubility limit
110 g > 100 g
Thus, the solution is saturated.
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Answer:
d = 0.93 g/cm³
Explanation:
Given data:
Mass of object = 28 g
Volume of object = 3cm×2cm×5cm
density of object = ?
Solution:
Volume of object = 3cm × 2cm ×5cm
Volume of object = 30 cm³
Density of object:
d = m/v
by putting values,
d = 28 g/ 30 cm³
d = 0.93 g/cm³