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WARRIOR [948]
3 years ago
13

What type of claim must be supported by scientifically valid evidence for it to be approved for use on a food label? A. Health c

laim B. Nutrient content claim C. Structure/function claim D. EER
Chemistry
2 answers:
Lelu [443]3 years ago
8 0
Option A. Health claim

Health claim must be supported by scientifically valid evidence to be approved in using  for food label.

This is also a food guide to all consumers the food content implied by manufacturers to their products.
nexus9112 [7]3 years ago
4 0

Answer is Health Claim

You might be interested in
5. Durante un estudio de la velocidad de la reacción A2(g) + 3B2(g)  2 AB3(g), se observa que en un recipiente cerrado que cont
weqwewe [10]

Answer:

a) Speed of the reaction = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time = 0.002083 mol/L.s

c) The rate of appearance of AB₃ = 0.004167 mol/L.s

Explanation:

English Translation

During a study of the reaction rate

A₂ (g) + 3B₂ (g) → 2 AB₃ (g),

it is observed that in a closed container containing a certain amount of A₂ and 0.75 mol / L of B₂, the concentration B₂ decreases to 0.5 mol / L in 40 seconds.

a) What is the speed of the reaction?

b) What is the rate of disappearance of A₂ during this period of time?

c) What is the rate of appearance of AB₃?

Solution

The rate of a chemical reaction is defined as the time rate at which a reactant is used up or the rate at which a product is formed.

It is the rate of change of the concentration of a reactant (rate of decrease of the concentration of the reactant) or a product (rate of increase in the concentration of the product) with time.

Mathematically, for a balanced reaction

aA → bB

Rate = -(1/a)(ΔA/Δt) = (1/b)(ΔB/Δt)

The minus sign attached to the change of the reactant's concentration indicates that the reactant's concentration decreases.

And the coefficients of each reactant and product in the balanced reaction normalize the rate of reaction for each of them

So, for our given reaction,

A₂ (g) + 3B₂ (g) → 2 AB₃ (g)

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

a) Speed of the reaction = Rate of the reaction

But we are given information on the change of concentration of B₂

Change in concentration of B₂ = ΔB₂ = 0.50 - 0.75 = -0.25 mol/L

Change in time = Δt = 40 - 0 = 40 s

(ΔB₂/Δt) = (-0.25/40) = -0.00625 mol/L.s

Rate of the reaction = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time

Recall

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

-(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

Rate of disappearance of A₂ = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

c) The rate of appearance of AB₃

Recall

Rate = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

(1/2)(ΔAB₃/Δt) = -(1/3)(ΔB₂/Δt)

(ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt)

rate of appearance of AB₃ = (ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt) = (-2/3) × (-0.00625) = 0.004167 mol/L.s

Hope this Helps!!!

3 0
3 years ago
The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0
3 years ago
The Earth produces a magnetic field in which the magnetic are most concentrated at the North and South
BlackZzzverrR [31]

Answer: I have no idea what kind of question this is, but I'm assuming it's a true or false question, and this is true. Although the north-bound pole is south-seeking, a compass would eventually draw you to the North Pole.

Explanation:

4 0
3 years ago
A 25.0 L tank of nitrogen gas is at 25 oC and 2.05 atm . If the temperature stays at 25 oC and the volume is decreased to 14.5 L
NeTakaya

Answer:

\boxed {\boxed {\sf P_2 \approx 3.53 \ atm}}

Explanation:

In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:

P_1V_1=P_2V_2

Now we can substitute any known values into the formula.

Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.

25.0 \ L * 2.05 \ atm = P_2V_2

The volume is decreased to 14.5 liters, but the pressure is unknown.

25.0 \ L * 2.05 \ atm = P_2 * 14.5 \ L

Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .

\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}=\frac{P_2 *14.5 \ L}{14.5 \ L}

\frac {25.0 \ L * 2.05 \ atm }{14.5 \ L}= P_2

The units of liters cancel.

\frac {25.0  * 2.05 \ atm }{14.5 }=P_2

\frac {50.25\  atm }{14.5 }=P_2

3.53448276 \ atm = P_2

The original values of volume and pressure have 3 significant figures, so our answer must have the same.

For the number we found, that is the hundredth place.

  • 3.53448276

The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.

3.53 \ atm \approx P_2

The new pressure is approximately <u>3.53 atmospheres.</u>

8 0
3 years ago
How many moles of methane (CH4) are in 7.31 x 10^25 molecules
Ivan

molar mass of methane CH4

= C + 4 H  

= 12.0 + 4 x 1.008

= 12.0 +  4.032

= 16.042g/mol

7.31 x 10^25 molecules x <u>             1 mole  CH4    </u>  = 121.43 moles

                                       6.02 x 10^23 CH4 molecules

121.43 moles CH4 are present.

       


6 0
3 years ago
Read 2 more answers
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