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2 years ago

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he chemical method of analysis in determination of the blood alcohol content (%BAC) is: K2Cr2O7 + H2SO4 + C2H5OH → Cr2(SO4)3 + K

2SO4 + CH3COOH + H2O What are the stoichiometric coefficients for the reaction above? (please enter the coefficients without spaces and with a coma between the numbers, e.g. 1,2,3,4)

1 answer:

emmasim [6.3K]2 years ago

3 0

Explanation:

Stoichiometric coefficients : these are the numeral which written before the chemical symbol of the chemical substance in a balanced chemical reaction.

We are given with the chemical equation which is used in blood alcohol content.The balanced chemical equation will be given as:

$2K_2Cr_2O_7 + 8H_2SO_4 + 3C_2H_5OH\rightarrow 2Cr_2(SO_4)_3 + 2K_2SO-4 + 3CH_3COOH + 11H_2O$

The stoichiometric coefficient of potassium dichromate = 2

The stoichiometric coefficient of sulfuric acid= 8

The stoichiometric coefficient of ethanol= 3

The stoichiometric coefficient of potassium sulfate = 2

The stoichiometric coefficient of chromium(III) sulfate = 2

The stoichiometric coefficient of acetic acid = 3

The stoichiometric coefficient of water = 11

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Carts A and B are identical and are moving toward each other on a track. The speed of cart A is v, while the speed of cart B is

borishaifa [10]

Answer: k= $\frac{5mv^{2} }{2}$

Explanation:

Recall that the formula for kinetic energy is given below as

k = $\frac{mv^{2} }{2}$

where k=kinetic energy (joules), m= mass of object (kg), v= velocity of object m/s)

For cart A

$m_{a}$ = mass of cart A

$v_{a}$ = v = velocity of cart A

$K.E_{a}$ = kinetic energy of cart A

hence, $K.E_{a}$ = $\frac{m_{a}v^{2} }{2}$

For cart B

$m_{b}$ = mass of cart B

$v_{b}$ = 2v = velocity of cart B

$K.E_{b}$ = kinetic energy of cart B

hence, $K.E_{b}$ = $\frac{m_{b}(2v^{2}) }{2}$ = 2 $m_{b} v^{2}$

from the question, both cart are identical which implies they have the same mass i.e $m_{a}$ = $m_{b}$ = m which implies that

$K.E_{a}= \frac{mv^{2} }{2}$ and $K.E_{b} =2mv^{2}$

The total kinetic energy K is the sum of cart A and cart B kinetic energy

$K=K.E_{a} + K.E_{b}$

$K=\frac{mv^{2} }{2} + 2mv^{2}$

hence

$K=\frac{5mv^{2} }{2}$

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2 years ago

You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon

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Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

$u_{1}$ = 0m/s

$t_{1}$ = t

$u_{2}$ = 43.12m/s

$t_{2}$ = (t-2.2)s

The distance by the first balloon is

$D = u_{1} t_{1} + \frac{1}{2} at_{1}^2$

where

a = 9.8m/s2

Inputting the values

$D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2$

The distance traveled by the second balloon

$D = u_{2} t_{2} + \frac{1}{2} at_{2}^2$

Inputting the values

$D = (43.12)(t-2.2) + \frac{1}{2} (9.8)(t-2.2)^2$

simplifying

$D = 4.9t^2 + 21.56t -71.148$

Substituting D of the first balloon into the D of the second balloon and solving

$4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s$

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

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7 0

2 years ago

For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat

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Answer:

Radius of the solenoid is 0.93 meters.

Explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation,

$B(t)=5t\ T$ , t is time in seconds

$\dfrac{dB}{dt}=5\ T$

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

$E(2\pi x)=\dfrac{d\phi}{dt}$

x is the distance from the axis of the solenoid

$E(2\pi x)=\pi r^2\dfrac{dB}{dt}$ , r is the radius of the solenoid

$r^2=\dfrac{2xE}{(dE/dt)}$

$r^2=\dfrac{2\times 2\times 1.1}{(5)}$

r = 0.93 meters

So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.

4 0

2 years ago

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A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its sta

serg [7]

Answer:

The final velocity of the bullet is 9 m/s.

Explanation:

We have,

Mass of a bullet is, m = 0.05 kg

Mass of wooden block is, M = 5 kg

Initial speed of bullet, v = 909 m/s

The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

$mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s$

So, the final velocity of the bullet is 9 m/s.

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2 years ago

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kkurt [141]

Answer:

$4.3\cdot 10^{-12} J$

Explanation:

The fusion reaction in this problem is

$4^1_1H \rightarrow ^4_2He +2e^+$

The total energy released in the fusion reaction is given by

$\Delta E = c^2 \Delta m$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\Delta m$ is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

$m(^1_1H)=1.007825u$ is the mass of one nucleus of hydrogen

$m(^4_2 He)=4.002603u$ is the mass of one nucleus of helium

So the mass defect is:

$\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u$

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$\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg$

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$\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J$

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2 years ago