Answer is: <span>the volume of water after the solid is added</span> is 4.5 ml.
d(gold) = 8.0 g/cm³; density of gold.
m(gold) = 4 g; mass of gold.
V(gold) = m(gold) ÷ d(gold); volume of gold.
V(gold) = 4 g ÷ 8 g/cm³.
V(gold) = 0.5 cm³ = 0.5 ml.
V(water) = 4.00 ml = 4.00 cm³.
V(flask) = V(gold) + V(water).
V(flask) = 0.5 cm³ + 4 cm³.V = 4.5 cm³.
It would be a gas because gases have the highest amount of kinetic energy and takes the shape/volume of it's container.
Hope this helps(;
Answer:
The electrophile is the hydroxide free radical
Explanation:
The hydroxylation of benzene and benzene derivatives using hydrogen peroxide proceeds in the presence of an acidic catalyst. The electrophile in this reaction is the hydroxyl free radical generated in an initial step of the reaction.
This is actually a free radical reaction. The hydroxyl radical previously generated reacts with the benzene ring to yield a radical that undergoes further rearrangement to yield the product phenol. The intermediate, shown as part of this reaction mechanism (refer to image attached) is a specie in which the odd electron is delocalized over the entire benzene ring. Loss of a proton completes the reaction mechanism yielding the corresponding phenol.
All matter (solid, liquid or gas) has a definite shape and occupies a specific volume.
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Hope it helps...</h2>
Answer:
11.3 g/mL.
Explanation:
- Density is a characteristic property of a substance.
- The density of a substance is the relationship between the mass of the substance and the volume it takes up.
<em>d = m/V,</em>
where, d is the density of the metal,
m is the mass of the metal (m = 12.475 g),
V is the volume that the metal takes up (V = 6.1 mL - 5.0 mL = 1.1 mL).
<em>∴ d = m/V =</em> (12.475 g)/(1.1 mL) = <em>11.34 g/mL ≅ 11.3 g/mL.</em>
<em>The significant figures has its rules in the calculations; the result has a significant figures to the lowest no. of significant figures (1.1 has two significant figures, while 12.475 has 5 significant figures), so the result should have only 2 significant figures.</em>
