Answer:
21 g/mL
Explanation:
To solve this problem, first look at the density equation, which is D=M/V, which D stands for density, M stands for mass, and V stands for volume. When you substitute in the variables, you get D=17.5/.82, which is equivalent to 21.34. However, since we need to pay attention to the sig fig rules for multiplying, we need to have the same amount of sig figs as the value with the least amount of sig figs, which is the number .82. .82 has two sig figs, so you round down. Your answer will be 21 g/mL.
Answer:
A: Synonyms
Explanation:
Synonyms are two words that have same meaning but are different words (tell me if this works!)
Answer:
4.2 g
Explanation:
The VOLUME of the ring is 4.2 - 4.0 = .2 ml = .2 cm^3
the MASS of the ring is this times the density
.2 cm^3 * 21 g/cm^3 = 4.2 g
Explanation:
The O atom is sp3 in a water molecule, with two sigma bonds and two lone pairs of electrons like that in water. The steric integer is thus 4, and its structure is tetrahedral.
The C atom is sp hybridised into two identical bonds and two identical bonds in acetylene.
The steric integer is therefore 2 because only sigma bonds are engaged in deciding hybridization, and its structure is linear.
The C atom is sp2 hybridised in ethene with single pi bond and three sigma identical bonds.
Thus the steric integer is 3, and its structure is planar trigonal.
The C atom is sp2 hybridized in ethene, with one pi bond and three sigma identical bonds.
The steric integer would therefore be 3 and its structure is planar trigonal.
The O atom is sp3 in a water molecule with two bond pairs and two lone pairs of electrons like that. The steric integer is thus 4, and its structure is tetrahedral.
The C atom is sp3 in a methane ring, with 4 bond pairs and no solitary pairs of electrons like that. The steric integer is thus 4, and its structure is tetrahedral.
Answer:
Explanation:
Temperature and thermal energy are in a direct proportion which means that if temperature of a substance increases, its thermal energy also increases and vice versa.
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Hope this helped!
<h3>~AH1807</h3>
