Answer:
350 g dye
0.705 mol
2.9 × 10⁴ L
Explanation:
The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:
70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye
The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:
350 g × (1 mol / 496.42 g) = 0.705 mol
The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:
3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L
Answer:
The reason why atomic mass is usually not a whole number is because it is a weighted average of the mass numbers of isotopes
Explanation:
Answer:
Zn =⇒ Zn+2(0.10) + 2e- (anode)
Zn+2(?M) + 2e- === Zn(s) (cathode)
Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn
E = E^o -0.0592 log Q; in this case E^o is zero.
E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2
23 mV x 1 volt/1000mv = 0.023 Volts
0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
0.023 = -0.0296 { log 0.10 – log [Zn+2] }
0.023 = -0.0296{ -1 - log[Zn+2] }
0.023 = +0.0296 + 0.0296log[Zn+2]
-0.0066 = 0.0296log[Zn+2]
-0.22= log[Zn+2]
[Zn+2] = 10^-0.22 = 0.603 Molar
Since the density of water is 1 g /mL, hence there is 100
g of H2O. So total mass is:
m = 100 g + 5 g = 105 g
=> The heat of reaction can be calculated using the
formula:
δhrxn = m C ΔT
where m is mass, C is heap capacity and ΔT is change in
temperature = negative since there is a decrease
δhrxn = 105 g * 4.18 J/g°C * (-2.30°C)
δhrxn = -1,009.47 J
=> However this is still in units of J, so calculate
the number of moles of NaCl.
moles NaCl = 5 g / (58.44 g / mol)
moles NaCl = 0.0856 mol
=> So the heat of reaction per mole is:
δhrxn = -1,009.47 J / 0.0856 mol
δhrxn = -11,798.69 J/mol = -11.8 kJ/mol
MThe heat energy required to raise the temperature of 0.36Kg of copper from 22 c to 60 c is calculate using the following formula
MC delta T
m(mass)= 0.360kg in grams = 0.360 x1000 = 360 g
c(specific heat energy) = 0.0920 cal/g.c
delta T = 60- 23 = 37 c
heat energy is therefore= 360g x0.0920 cal/g.c x 37 c= 1225.44 cal
