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3 years ago

If we eat a sandwich, why does it not come out as a

Chemistry

2 answers:

xxTIMURxx [149]3 years ago

7 0

Answer:

Is this something logical? I don't know but it sounds absurd.

Sunny_sXe [5.5K]3 years ago

4 0

Answer:

if we eat a sandwich it doesn't come out as a sandwich because we already ate it. its not going to come out whole if we already chewed it and digested it-

Explanation:

hope this helps, ỉdk if this is a real question, but if it is, then i guess use this answer??

You might be interested in

Which statement about the balanced equations for nuclear and chemical changes is correct? (1 point)

iris [78.8K]

The true statement about the balanced equations for nuclear and chemical changes is; both are balanced according to the total mass before and after the change.

A basic law in science is called the law of conservation of mass. Its general statement is that mass can neither be created nor destroyed.

Both in chemical and nuclear changes, mass is involved and in both cases, the law of conservation of mass strictly applies.

This means that for both chemical and nuclear changes; total mass before reaction must be equal to total mass after reaction.

Hence, both reactions are balanced according to the total mass before and after the change.

Learn more: brainly.com/question/22064431

3 0

2 years ago

Read 2 more answers

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

Answer: The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

3 years ago

A student obtains a mixture of the chlorides of two unknown metals, X and Z. The percent by mass of X and the percent by mass of

Aleksandr-060686 [28]

Answer: The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

Explanation:

To calculate the mole percent of a substance, we use the equation:

$\text{Mole percent of a substance}=\frac{\text{Moles of a substance}}{\text{Total moles}}\times 100$

Mass percent means that the mass of a substance is present in 100 grams of mixture

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively

Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X

8 0

3 years ago

Calculate the molarity of a solution of Nach if it contains 7.2.g Nach in 100.0 mL of solution. andver: m Nach . .

Norma-Jean [14]

Answer:

1.23 M

Explanation:

Molarity of a substance , is the number of moles present in a liter of solution .

M = n / V

M = molarity

V = volume of solution in liter ,

n = moles of solute ,

Moles is denoted by given mass divided by the molecular mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molecular mass .

From the question ,

w = given mass of NaCl = 7.2 g

As we know , the molecular mass of NaCl = 58.5 g/mol

Moles is calculated as -

n = w / m = 7.2 g / 58.5 g/mol = 0.123 mol

Molarity is calculated as -

V = 100ml = 0.1 L (since , 1 ml = 1/1000L )

M = n / V = 0.123 mol / 0.1 L = 1.23 M

5 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