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3 years ago

Which of these is a characteristic of science?

Chemistry

2 answers:

jenyasd209 [6]3 years ago

5 0

The correct answer is D. It gives the same result when experiments are repeated.

Explanation:

Science refers to a system or discipline that aims at understanding phenomena that is testable mainly through multiple observations, hypotheses, and testing. Because of this, characteristics of science include that it is based on facts or ideas that have been proved, it is useful for answering testable questions and also the findings are replicable which means if tested multiple times same results would be obtained which proves phenomenons and events are facts and do not depend on opinions or personal views. Considering this, the one that is a characteristic of science is that "It gives the same result when experiments are repeated".

Ivahew [28]3 years ago

3 0

D. It gives the same results when experiments are repeated

You might be interested in

HClO4+P4O10=H3PO4+Cl2O7 how do you balance this?

Tom [10]

12 HClO₄ + 1 P₄O₁₀ → 4 H₃PO₄ + 6 Cl₂O₇

<h3>Explanation</h3>

Balance by the conservation of atoms.

Assign coefficient 1 to the species with the largest number of elements and atoms. H₃PO₄ contains three elements. Each of its molecule contains eight atoms, that's two more than the six atoms in a HClO₄ molecule. Start by assigning H₃PO₄ a coefficient of 1.

? HClO₄ + ? P₄O₁₀ → 1 H₃PO₄ + ? Cl₂O₇

There are now three H atoms, one P atom on the product side. H₃PO₄ is the only product that contains H and P atoms. As a result, there should be the same number of H and P atoms on the reactant side.

Among all reactants, only HClO₄ contains H atoms. Each HClO₄ molecule contains one H atom. Three H atoms correspond to three HClO₄ molecule.
Among all reactants, only P₄O₁₀ contains P atoms. Each P₄O₁₀ molecule contains four P atoms. One P atom corresponds to 1/4 of a P₄O₁₀ molecule.

Thus

3 HClO₄ + 1/4 P₄O₁₀ → 1 H₃PO₄ + ? Cl₂O₇

There are three Cl atoms in three HClO₄ molecules. HClO₄ is the only species that contains Cl among all reactants. There are three Cl atoms on the reactant side and shall be the same number of Cl atoms on the product side.

Cl₂O₇ is the only molecule that contains Cl among the products. Each Cl₂O₇ molecule contains two Cl atoms. Three Cl atoms will correspond to 3/2 Cl₂O₇ molecules.

3 HClO₄ + 1/4 P₄O₁₀ → 1 H₃PO₄ + 3/2 Cl₂O₇

Multiply both sides by the least common multiple of the denominators to eliminate the fraction. The least common multiple in this case is four.

12 HClO₄ + 1 P₄O₁₀ → 4 H₃PO₄ + 6 Cl₂O₇

5 0

2 years ago

A chemist titrates 190.0 mL of a 0.8125 M ammonia (NH) solution with 0.3733 M HCl solution at 25 °C. Calculate the pH at equival

stealth61 [152]

Answer:

Approximately $4.92$ .

Explanation:

Initial volume of the solution: $V = 190.0\; \rm mL = 0.1900\; \rm L$ .

Initial quantity of $\rm NH_3$ :

$\begin{aligned} n({\rm NH_3}) &= c({\rm NH_3}) \cdot V({\rm NH_3}) \\ &= 0.3733\; \rm mol \cdot L^{-1} \times 0.1900\; \rm L \\ &\approx 0.154375\; \rm mol\end{aligned}$ .

Ammonia $\rm NH_3$ reacts with hydrochloric $\rm HCl$ acid at a one-to-one ratio:

$\rm NH_3 + HCl \to NH_4 Cl$ .

Hence, approximately $n({\rm HCl}) = 0.154375\; \rm mol$ of $\rm HCl\!$ molecules would be required to exactly react with the $\rm NH_3\!$ in the original solution and hence reach the equivalence point of this titration.

Calculate the volume of that $0.3733\; \rm mol \cdot L^{-1}$ $\rm HCl$ solution required for reaching the equivalence point of this titration:

$\begin{aligned}V({\rm HCl}) &= \frac{n({\rm HCl})}{c({\rm HCl})} \\ &\approx \frac{0.154375\; \rm mol}{0.3733\; \rm mol \cdot L^{-1}} \approx 0.413541\; \rm L\end{aligned}$ .

Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately $0.413541\; \rm L + 0.1900\; \rm L \approx 0.6035\; \rm L$ .

If no hydrolysis took place, $0.154375\; \rm mol$ of $\rm NH_4 Cl$ would be produced. Because $\rm NH_4 Cl\!$ is a soluble salt, the solution would contain $0.154375\; \rm mol\!$ of $\rm {NH_4}^{+}$ ions. The concentration of $\rm {NH_4}^{+}\!$ would be approximately:

$\begin{aligned}c({\rm {NH_4}^{+}}) &= \frac{n({\rm {NH_4}^{+}})}{V({\rm {NH_4}^{+}})}\\ &\approx \frac{0.154375\; \rm mol}{0.6035\; \rm L} \approx 0.255782\; \rm mol \cdot L^{-1}\end{aligned}$ .

However, because $\rm NH_3 \cdot H_2O$ is a weak base, its conjugate $\rm {NH_4}^{+}$ would be a weak base.

$\begin{aligned}pK_{\rm a}({{\rm NH_4}}^{+}) &= pK_{\rm w} - pK_{\rm b}({\rm NH_3})\\ &\approx 13.99 - 4.75 = 9.25\end{aligned}$ .

Hence, the following reversible reaction would be take place in the solution at the equivalence point:

$\rm {NH_4}^{+} \rightleftharpoons NH_3 + H^{+}$ .

Let $x\; \rm mol \cdot L^{-1}$ be the increase in the concentration of $\rm H^{+}$ in this solution because of this reversible reaction. (Notice that $x \ge 0$ .) Construct the following $\text{RICE}$ table:

$\begin{array}{c|ccccc} \textbf{R}& \rm {\rm NH_4}^{+} & \rightleftharpoons & {\rm NH_3}& + & {\rm H}^{+}\\ \textbf{I} & 0.255782 \; \rm M \\ \textbf{C} & -x \;\rm M & & + x\;\rm M & & + x\; \rm M \\ \textbf{E} & (0.255782 - x)\; \rm M & & x\; \rm M & & x\; \rm M\end{array}$ .

Thus, at equilibrium:

Concentration of the weak acid: $[{\rm {NH_4}^{+}}] \approx (0.255782 - x) \; \rm M$ .
Concentration of the conjugate of the weak acid: $[{\rm NH_3}] = x\; \rm M$ .
Concentration of $\rm H^{+}$ : $[{\rm {H}^{+}}] \approx x\; \rm M$ .

$\displaystyle \frac{[{\rm NH_3}] \cdot [{\rm H^{+}}]}{[{ \rm {NH_4}^{+}}]} = 10^{pK_\text{a}({\rm {NH_4}^{+}})}$ .

$\displaystyle \frac{x^2}{0.255782 - x} \approx 10^{-9.25}$

Solve for $x$ . (Notice that the value of $x\!$ is likely to be much smaller than $0.255782$ . Hence, the denominator on the left-hand side $(0.255782 - x) \approx 0.255782$ .)