Answer:
<h2>The answer is 7.14 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula

From the question
mass of metal = 25 g
volume = final volume of water - initial volume of water
volume = 28.5 - 25 = 3.5 mL
It's density is

We have the final answer as
<h3>7.14 g/mL</h3>
Hope this helps you
Answer : The correct answer for a) 4-bromo-2-iodo-4-methyl pentane and b)5-bromo-2-ethoxy-2-methyl pentane.
A) Reaction with NaI :
Reaction of alkyl halide with NaI is known as Finkelstein Reaction . The acetone is used as solvent . It involves bimolecular nucleophillic substitution rmechanism (SN²) . There is replecement of one halogen with other occurs .
The incoming Nucleophile(Nu⁻) (halide) attacks on carbon from back side , while the leaving group (halide) leaves the compound from front side , simultaneously. The product so formed have is inverted .(Image)
NaI releases I⁻ ion which act as nucelophile and attacks on C1 carbon and Br⁻ from C1 carbon is released . Out of two bromines at C1 and C4 carbons , C1 is primary carbon which is less sterically hindered while C-4 is tertiary carbon and sterically hindered . So it is easy for incoming Nu⁻ to attack on C1 carbon .So Br⁻ is repleaced by I⁻.
1,4-dibromo-4-methylpentane + NaI → 4-bromo-1-iodo-4-methylpentane
The product formed from reaction between 1,4-dibromo-4-methylpentane and NaI is 4-bromo-1-iodo-4-methylpentane . (Image)
B) Reaction with AgNO3 :
Reaction of alkyl halide with AgNO3 in ethanol takes place via SN¹ ( unimolecular nucleophilic substitution ) mechanism . In this leaving group(halide) leaves from alkyl halide forming an intermediate carbocation species . The incoming Nu⁻ attack on this carbocation.
AgNO3 reacts releases Ag⁺ion which abstract Br⁻ of C-4 carbon from 1,4-dibromo-4-methylpentane. THis forms tertiary carbocation which is more stable than carbocation formed by removal of Br from C-1 . The ethanol being more Nucleophilic than NO₃⁻ (from AgNO₃), attacks on this carbocation .(Image )
The product formed as a result is 5-bromo-2-ethoxy-2-methyl pentane.
Answer:
Yes, because the light was the manipulated variable
Explanation:
The one that is observed or measured in the experiment, and it is known as the dependent variable.
1. 1.636 moles
2. 271.06 kPa pressure
3. Tires will be burst
4. 235.91 kPa
Explanation:
Step 1:
PV = nRT, is the equation to be used where
P represents pressure
V represents volume
n represents moles of gas
R is constant
T represents temperature in Kelvin
n=RT/PV
It is given that the pressure is 245 kPa at initial temperature 19 C and tire volume is 16.2 L. Temperature must be converted to Kelvin, 19 C equals 292K.
n=PV/RT ->245*16.2/(8.31*292) = 1.636
Number of moles of Nitrogen in the tire = 1.636
Step 2:
We need to find the maximum tire pressure at 50 C (323K)
P = nRT/V
Substituting the values P = (1.636 * 8.31 * 323)/16.2 = 271.06 kPa
The tire pressure at 50 C will be 271.06 kPa
Step 3
We need to figure out if the tires would burst in Chelan when the temperature is 55 C. It is given that the maximum pressure the tires can withstand is 265 kPA, so any pressure above this will cause the tire to burst. In Step-2 we calculated that the pressure is 271.06 kPA at 50 C which is more than the maximum pressure the tire can withstand. The pressure would increase further with temperature and at 55 C the pressure will be more than 271.06 kPa. So the tires are likely to burst in Chelan.
Step 4:
We need to find the pressure of Nitrogen at 19 C before the start of the trip so that tires will not burst. The pressure at 55 C is 265 kPa. Let us find the number of moles at this temperature and pressure.
n= PV/RT -> n=265*16.2/(8.31*328) = 1.575
Now let us find the pressure at 19 C.
P = nRT/V -> 1.575*8.31*292/16.2 = 235.91 kPa