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olga nikolaevna [1]
3 years ago
6

If the caffeine concentration in a particular brand of soda is 1.87 mg/oz, drinking how many cans of soda would be lethal? Assum

e 10.0 grams of caffeine is a lethal dose, and there are 12 oz in a can.
Chemistry
1 answer:
Drupady [299]3 years ago
4 0
2.77mg caffeine / 1oz12oz / 1canLethal dose: 10.0g caffeine = 10,000mg caffeine First, find how much caffeine is in one can of soda, then divide that amount by the lethal dose to find the number of cans. (2.77mg caffeine / 1oz) * (12oz / 1can) = 33.24mg caffeine / 1can. (10,000mg caffeine) * (1can / 33.24mg caffeine) = 300.84 cans. Since we can't buy parts of a can of soda, then we have to round up to 301 cans. Notice how all the values were set up as ratios and how the units cancelled.
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Which change does an increase in heat energy cause?
Bumek [7]

Answer:

C more collision between molecules

Explanation:

increase in heat causes increase in kinetic energy of the particles

5 0
2 years ago
The lowest whole-number ratio of elements in a compound is called the ?
ASHA 777 [7]

Answer:

The Empirical Formula.

Explanation:

From the empirical formula and using the weight (in g)  of a given substance, we can come up with the molecular formula which is the actual weight of a substance. Sometimes, we find that the empircal formula is the molecular formula.

8 0
3 years ago
Under certain conditions, the equilibrium constant of the reaction below is Kc=1.7×10−3. If the reaction begins with a concentra
svetoff [14.1K]

Answer:

[Cl2] equilibrium = 0.0089 M

Explanation:

<u>Given:</u>

[SbCl5] = 0 M

[SbCl3] = [Cl2] = 0.0546 M

Kc = 1.7*10^-3

<u>To determine:</u>

The equilibrium concentration of Cl2

<u>Calculation:</u>

Set-up an ICE table for the given reaction:

               SbCl5(g)\rightleftharpoons SbCl3(g)+Cl2(g)

I                 0                    0.0546     0.0546

C              +x                        -x               -x

E               x                  (0.0546-x)    (0.0546-x)

Kc = \frac{[SbCl3][Cl2]}{[SbCl5]}\\\\1.7*10^{-3} =\frac{(0.0546-x)^{2} }{x} \\\\x = 0.0457 M

The equilibrium concentration of Cl2 is:

= 0.0546-x = 0.0546-0.0457 = 0.0089 M

5 0
3 years ago
Read 2 more answers
The ka for hcn is 4.9 ⋅ 10-10. What is the value of kb for cn-?
marusya05 [52]

Answer:

kb = 2,0x10⁻⁵

Explanation:

The ka for HCN is:

HCN ⇄ H⁺ + CN⁻; ka = 4,9x10⁻¹⁰ <em>(1)</em>

The inverse reaction has an equilibrium constant of:

H⁺ + CN⁻ ⇄ HCN k = 1/4,9x10⁻¹⁰ = 2,0x10⁹ <em>(2)</em>

As the equilibrium of the water is:

H₂O ⇄ H⁺ + OH⁻; kw = 1x10⁻¹⁴ <em>(3)</em>

The sum of (2) + (3) gives:

H₂O + CN⁻ ⇄ HCN + OH⁻; kb = kw×k = 1x10⁻¹⁴×2,0x10⁹ =

2,0x10⁻⁶; <em>kb = 2,0x10⁻⁵</em>

<em />

<em>-In fact, the general formula to convert from ka to kb is:</em>

<em>kb = kw / ka-</em>

<em />

I hope it helps!

5 0
3 years ago
How would you know if a precipitate was formed without performing the chemical reaction
Goshia [24]
Hey there

Thats easy

Precipitation occurs when cations and anions in aqueous solution combine to form an insoluble ionic compound called precipitation. So, you can refer to a solubility chart or draw polar/nonpolar lewis structures, that might be helpful.
Remember water is polar so polar molecules will dissolve in water 
4 0
3 years ago
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