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3 years ago

In chemical formulas, there are often subscript numbers (such as the 2 in H2O, or the 6, 12, and 6 in C6H12O6).

1 answer:

8 0

Answer : The correct option is, Subscript numbers tell us how many of the element to their upper-left there are in the molecule.

Explanation :

Subscript number : It indicates that the total number of atoms of an element present in a molecule.

While witting a chemical formula, first we have to write the symbol of an element and then write the number in the bottom of the symbol.

For example :

$C_6H_{12}O_6$ is the chemical formula of the glucose molecule. The glucose molecule is made up of six carbon atoms, twelve hydrogen atoms and 6 oxygen atoms.

The number 6 of carbon atom is written after and in the bottom (lower-right) of the symbol of carbon element (C). Similarly, the number 12 is written lower-right of the symbol of hydrogen element (H) and the number 6 is written lower-right of the symbol of oxygen element (O).

Therefore, the subscript numbers tell us how many of the element to their upper-left there are in the molecule.

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D. nuclear fusion because atomic nuclei combine to form a heavier nucleus

Explanation:

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2 years ago

Read 2 more answers

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The followin

erastovalidia [21]

Answer : The initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

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:

$A+B+C\rightarrow D+E$

Rate law expression for the reaction:

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

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.0\times 10^{-5}=k(0.20)^a(0.20)^b(0.20)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.20)^a(0.20)^b(0.60)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.40)^a(0.20)^b(0.20)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.40)^a(0.40)^b(0.20)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.0\times 10^{-5}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

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

Putting values in equation 1, we get:

$6.0\times 10^{-5}=k(0.20)^2(0.20)^0(0.20)^1$

$k=7.5\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.75 M of reagent A and 0.90 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.5\times 10^{-3})\times (0.75)^2(0.90)^0(0.90)^1$

$\text{Rate}=3.4\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.4\times 10^{-3}Ms^{-1}$

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2 years ago

When compounds are formed, will its heat give off?

mariarad [96]

Answer: It depends on the type of chemical reaction that formed the compound.

Explanation:

Exothermic reactions give off the heat to the reaction environment, so the compound feels hotter.

Endothermic reactions absorb the heat from the reaction environment and the compound feels cooler.

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3 years ago

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Answer:

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2 years ago

Unknown # 41

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Answer: Potassium Iodide, KI

Explanation:

Flame test colors:

Li+ = Crimson Red
Na+ = Bright Orange-Yellow
K+ = Lilac

Addition of nitric acid and silver nitrate (HNO3 and AgNO3),

Cl- = White precipitate
Br- = Creamy precipitate
I- = Yellow Precipitate

Hope this helps, brainliest would be appreciated :)

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3 years ago