Answer:
C. 1 cubic foot of loose sand
Explanation:
For many objects having equal volume , surface area will be maximum
of the object which has spherical shape .
But when a sphere is broken into tiny small spheres , total surface area of all the small spheres will be more than surface area of big sphere .
Hence among the given option , surface area of loose sand will have greatest surface area . Loose sand is equivalent to small spheres .
Hey there!
Static electricity is a missing charge. When positive ions are formed (where an electron is missing), this imbalance is called static electricity.
Answer:
A. SO4 2−
Explanation:
SO₄²⁻
A covalent bond is a bond that forms from sharing of electrons between two atoms. Here, the two atoms combined must have a very low electronegative difference between them, usually 0. Electronegativity deals with the tendency of an atom to attract electrons to itself.
Oxygen and sulfur shows similar tendencies. Since the two atoms shows a strong affinity for their valence electrons in order to complete the octet. This will lead to the eventual sharing of the valence electrons.
Answer:
0.80m of KOH
Explanation:
Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.
In the problem, the solute is KOH and solvent is water.
Moles of 36g KOH -Molar mass: 56.1g/mol- are:
36g KOH × (1mol / 56.1g) = <em>0.642 moles of KOH</em>
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Now, as density of water is 1g/mL, mass of 800mL of water is:
800mL × (1g / mL) × (1kg / 1000g) = <em>0.800kg of water</em>
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Thus, molality is:
0.642moles of KOH / 0.800kg = <em>0.80m of KOH</em>
The volume of the NaOH used is calculated as 14 mL.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with the calculation of the amount of substance in a reaction using mass - mole or mass - volume relationship.
Here;
Number of moles of CaCO3 = 0.205 g/100.1 = 0.00205 moles
Number of moles of HCl = 2.00 M * 7/1000 L = 0.014 moles
2 moles of HCl reacts with 1 mole of CaCO3
x moles of HCl reacts with 0.00205 moles of CaCO3
x = 0.00205 moles * 2/1 = 0.0041 moles
Hence HCl is the excess reactant
Amount of excess HCl = 0.014 moles - 0.0041 moles = 0.0099 moles
Concentration of excess HCl reacted = 0.0099 moles/125 * 10^-3 = 0.0792 M
Using;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
VB = CAVANB/CBNA
VB = 0.0792 M * 10 mL * 1/ 0.058 M
VB = 14 mL
Missing parts;
A 0.205 g sample of caco3 (mr = 100.1 g/mol) is added to a flask along with 7.50 ml of 2.00 m hcl. caco3(aq) + 2hcl(aq) → cacl2(aq) + h2o(l) + co2(g) enough water is then added to make a 125.0 ml solution. a 10.00 ml aliquot of this solution is taken and titrated with 0.058 m naoh. naoh(aq) + hcl(aq) → h2o(l) + nacl(aq) how many ml of naoh are used?
Learn more about stoichiometry: brainly.com/question/9743981
