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1 year ago

A substance is uniform in its composition. which term best defines this substance?

Chemistry

1 answer:

Nady [450]1 year ago

5 0

The term that best describes the substance is uniform in its composition is pure substance.

<h3>What are pure substance?</h3>

A pure substance is a substance that is made up of only one kind of particle and have a fixed or constant composition.

Pure substances can be further classified as either;

Elements; any one of the simplest chemical substances that cannot be decomposed in a chemical reaction or by any chemical means and made up of atoms.

Compounds; a pure substance formed by chemical union of two or more ingredients in definite proportions by weight.

Therefore, the term that best describes the substance is uniform in its composition is pure substance.

Learn more about pure substance at: brainly.com/question/24462192

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How many moles of na2so4 are present in 284.078 grams of na2so4? a. 2 b. 3 c. 5 d. 6

schepotkina [342]

There are 2 moles of NA2SO4 present in 284.078 grams of NA2SO4. Why? Well, we already know here that there are 284.078 grams of NA2SO4. To find how many moles are present in this compound, we would have to divide these grams by the molar mass of the compound. When we go to the periodic table of elements, we will find that there are two moles sodium (which represents 45.98 grams), one mole sulfur (which represents 32.06 grams), and four moles oxygen (which represents 64 grams). When combining these, we would get one mole of sodium sulfate which is representative of 142.04 grams total. That is our molar mass. Now that we have the molar mass, we can go back and divide. 284.078 divided by 142.04 is equal to about 1.99999, and that rounds up to 2 moles. Hence, there are 2 moles of NA2SO4 present in 284.078 grams of NA2SO4.

Your final answer: A is your answer.

3 0

2 years ago

A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached

Rashid [163]

Answer:

(a) Cu²⁺ +2e⁻ ⇌ Cu

Explanation:

(a) Cu half-reaction

Cu²⁺ + 2e⁻ ⇌ Cu

(c) Cell voltage

The standard reduction potentials for the half-reactions are+

E°/V

Cu²⁺ + 2e⁻ ⇌ Cu; 0.34

Hg₂Cl₂ + 2e⁻ ⇌ 2Hg + 2Cl⁻; 0.241

The equation for the cell reaction is

E°/V

Cu²⁺(0.1 mol·L⁻¹) + 2e⁻ ⇌ Cu; 0.34

2Hg + 2Cl⁻ ⇌ Hg₂Cl₂ + 2e⁻; -0.241

Cu²⁺(0.1 mol·L⁻¹) + 2Hg + 2Cl⁻ ⇌ Cu + Hg₂Cl₂; 0.10

The concentration is not 1 mol·L⁻¹, so we must use the Nernst equation

(ii) Calculations:

T = 25 + 273.15 = 298.15 K

$Q = \dfrac{\text{[Cl}^{-}]^{2}}{ \text{[Cu}^{2+}]} = \dfrac{1}{0.1} = 10\\\\E = 0.10 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(10)\\\\=0.010 -0.01285 \times 2.3 = 0.10 - 0.03 = \textbf{0.07 V}\\\text{The cell potential is }\large\boxed{\textbf{0.07 V}}$

3 0

3 years ago

20. Which of the following is a method of purifying water? A. Adding iodine B. Freezing C. Aeration D. Fixation

Oxana [17]

The answer is A, adding Iodine

7 0

3 years ago

Read 2 more answers

A chemistry student needs 65.0 mL of carbon tetrachloride for an experiment. By consulting the CRC Handbook of Chemistry and Phy

Anna007 [38]

Answer:

103.3 g

Explanation:

Data Given:

Volume of carbon tetrachloride (v) = 65.0 mL

Density of carbon tetrachloride (d) = 1.59 g/cm³

Solution:

Convert volume of carbon tetrachloride from ml to cm³

1 ml = 1 cm³

so,

65.0 mL = 65.0 cm³

Formula used to calculated mass of carbon tetrachloride should be taken for experiment.

d = m/v

Rearrange the above equation:

m = d x v . . . . . . .(1)

Put values in equation 1

m = 1.59 g/cm³ x 65.0 cm³

m = 103.3 g

6 0

4 years ago

Which of the following is an oxidation-reduction reaction? a. HCl(aq) + LiOH(aq) → LiCl(aq) + H2O(l) b. Pb(C2H3O2)2(aq) + 2 NaCl

saw5 [17]

Answer: Option (d) is the correct answer.

Explanation:

A reaction in which there occurs change in oxidation state of reacting species is known as an oxidation-reduction reaction.

(a) $HCl(aq) + LiOH(aq) \rightarrow LiCl(aq) + H_{2}O(l)$

Will be written as:

$H^{+} + Cl^{-} + Li^{+} + OH^{-} \rightarrow Li^{+} + Cl^{-} + H^{+} + OH^{-}$

In this reaction, there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(b) $Pb(C_{2}H_{3}O_{2})_{2}(aq) + 2NaCl(aq) \rightarrow PbCl_{2}(s) + 2 NaC_{2}H_{3}O_{2}(aq)$

Will be written as:

$Pb^{2+} + 2C_{2}H_{3}OO^{-} + 2Na^{+} + 2Cl^{-} \rightarrow Pb^{2+} + 2Cl^{-} + 2Na^{+} + 2C2H3OO^{-}$

Similarly here, there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

Will be written as:

$Na^{+} + I^{-} + Ag^{2+} + NO^{2-}_{3} \rightarrow AgI(s) + Na^{+} + NO^{2-}_{3}(aq)$

Here, also there occurs no change in oxidation state of reacting species. Hence, it is not an oxidation-reduction reaction.

(d) $Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)$

So, here there occurs change in oxidation state of Mg from 0 to +2 and oxidation state of H changes from +1 to 0. Hence, it is an oxidation-reduction reaction.

Thus, we can conclude that $Mg(s) + 2 HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g)$ is an oxidation-reduction reaction.

7 0

4 years ago