NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol
<span>Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27% </span>
<span>3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3
</span><span>0.8211 grams Na + 1.266 grams Cl = 2.087 grams</span>
Answer:
Size and Temperature or E & B
Explanation:
Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive. Since soft nucleophiles are less strongly solvated than hard nucleophiles, these solvents boost the relative reactivity of soft anions.
<h3>
Ethanol is either a nucleophile or a base.</h3>
The ethanol is a base Because carbocation is an extremely reactive species, a base or nucleophile as weak as ethanol can replace or remove it. SN1 and E1 would not be conceivable without the carbocation or a strong departing group.
<h3>How do solvents impact anionic nucleophile's reactivity?</h3>
In polar aprotic solvents, nucleophilic substitution reactions of anionic nucleophiles often proceed more quickly. The normal relative reactivity order in such solvents (like DMSO)is Anions are solvated in protic hydrogen-bonding solvents (such as ethanol). Consequently, nucleophiles are less reactive.
Learn more about nucleophiles here:-
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I assume what you're asking about is, how does the temperature changes when we increase water's mass, according the formula for heat ?
Well the formula is :
(where Q is heat, m is mass, c is specific heat and
is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless, if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature
Answer: 15062.4 Joules
Explanation:
The quantity of heat energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = ?
Mass of food = 200.0g
C = 4.184 j/g°C
Φ = (Final temperature - Initial temperature)
= 83.0°C - 65.0°C = 18°C
Then, Q = MCΦ
Q = 200.0g x 4.184 j/g°C x 18°C
Q = 15062.4 J
Thus, 15062.4 joules of heat energy was contained in the food.
