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fenix001
12 days ago
14

Why couldn't you substitute 3m H2SO4 for concentrated HNO3 when oxidizing copper?

Chemistry
2 answers:
lorasvet [2.7K]12 days ago
5 0
Each oxidizing agent exhibits unique characteristics tailored to various reactants. Strength varies across different solutions of oxidizing agents. For instance, sulfuric acid (H2SO4) is not as potent as nitric acid (HNO3); therefore, utilizing H2SO4 to oxidize copper is ineffective since copper requires a more robust oxidizing agent.
Alekssandra [3K]12 days ago
5 0

This appears to be an essay question based on an experiment. Different oxidizing agents possess varying strengths. When compared to nitric acid (HNO3), sulfuric acid (H2SO4) serves as a weaker oxidizing agent. Therefore, using H2SO4 in place of HNO3 would not yield the same experimental results.



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On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.
KiRa [2933]

Answer:

  • Please refer to the attached image for the graph with labeled axes and points.

Explanation:

This is a guide to fulfilling the instructions along with essential notes for understanding how to create such graphs:

1) The horizontal axis should indicate pressure ranging from 0 mb to 760 mb, while the vertical axis corresponds to volume ranging from 0 to 1 mL.

The x-axis captures the independent variable, and the y-axis records the dependent variable. Both axes must be accurately labeled, showing the variable names and their respective units.

In this context, the origin, (0,0), signifies the intersection of the axes at a pressure of 0 mb and a volume of 0.0 milliliters.

2) Allocate values for the divisions on the axes to maximize the usage of space on both.

An effective graph aims to utilize the entire space on both axes; for this, identify the maximum values for pressure and volume, and determine the corresponding marks.

The pressure range along the x-axis is [90, 760 mb], suggesting large divisions of 100 mb, with the farthest right mark at 800 mb. You can then subdivide each 100 mb interval into 10 smaller sections, using small divisions of 10 mb (my example employs 4 sections of 25 mb, but 10 mb is preferable).

The volume's range for the vertical axis is [0.1, 0.8], so it’s best to use divisions set at 0.1 ml.

3) Next, identify and label the points as follows:

  • (90, 0.9) ⇒ 90 mb, 0.9 ml
  • (100, 0.8) ⇒ 100 mb, 0.8 ml
  • (400, 0.2) ⇒ 400 mb, 0.2 ml
  • (600, 0.15) ⇒ 600 mb, 0.15 ml
  • (760, 0.1) ⇒ 760 mb, 0.1 ml

The points represented as (x, y) are referred to as ordered pairs, indicating that the sequence is significant: the first number denotes the independent variable whereas the second denotes the dependent variable.

Thus, for the point (90, 0.9), 90 indicates a pressure of 90 mb and 0.9 indicates a volume of 0.9 ml.

To find (600, 0.15), since the horizontal increments are valued at 0.1, you should place the second coordinate of the point between the marks corresponding to 0.1 and 0.2 ml.

This allows you to accurately plot each point on the graph.

5 0
20 days ago
Read 2 more answers
Put the following elements into five pairs of elements that have similar chemical reactivity: F, Sr, P, Ca, O, Br, Rb, Sb, Li, S
Anarel [2989]

Explanation:

Elements provided:

  F, Sr, P, Ca, O, Br, Rb, Sb, Li, S

Elements sharing similar reactivity belong to the same group in the periodic table, indicating that those in the same column exhibit comparable reactivity. Here are the identified groupings:

  Li and Rb are alkali metals in group 1

  Ca and Sr are alkaline earth metals in group 2

  F and Br are halogens in group 7

  O and S belong to group 6

 P and Sb are classified in group 5 of the periodic table

Thus, these classifications illustrate elements with the same chemical characteristics.

7 0
1 month ago
If your lungs were filled with air containing this level of lead, how many lead atoms would be in your lungs? (Assume a total lu
Tems11 [2777]
A total of 1.505×10^23 lead atoms

In the lungs, the volume of lead equals the total lung volume, which is 5.60L

1 mole corresponds to 22.4L

Thus, 5.6L of lead converts to 5.6/22.4 = 0.25 moles

According to Avogadro's law

1 mole of lead contains 6.02×10^23 lead atoms

Thus, 0.25 moles of lead equates to 0.25×6.02×10^23 = 1.505×10^23 lead atoms

6 0
14 days ago
Does the result of the calculation in question 3 justify your original assumption that all of the SCN^- is in the form of FeNCS^
lorasvet [2795]
Fe 3+ + SCN- --> FeSCN 2+ 

<span>.......Fe 3+.......SCN-.........FeSCN 2+ </span>
<span>I.......0.04..........0.001.............. </span>
<span>C........-x...............-x............. </span>
<span>E.....0.04-x.....0.001-x...........x </span>

<span>Keq = 203.4 = x / (0.04-x)(0.001-x) </span>
<span>203.4 = x / (x^2 - 0.041x + 4x10^-5) </span>
<span>203.4x^2 - 8.34x + 0.00094 = x </span>
<span>203.4x^2 - 9.34x + 0.00094 = 0 </span>
<span>x = -0.0001M or 0.0458M </span>
<span>therefore, according to the calculated Keq, all of the SCN- and Fe 3+ would be fully converted into FeSCN 2+</span>
5 0
1 month ago
Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter
KiRa [2933]

Answer:

thickness is 0.29 cm

Explanation:

To create a fake iron ball out of gold, we must ensure that its mass matches that of the iron ball. Therefore, we first find the volume of the iron ball using the provided diameter, applying the formula of 4/3 pi r^3.

Given the diameter d = 6 cm; thus, the radius r = 3 cm (d/2).

We calculate the volume of the iron ball: 4/3 * 3.14 * 3^3 = 113.04 cm^3.

The corresponding mass of the iron ball is the volume multiplied by its density: 113.04 * 5.15 g/cm^3 = 582.156 g.

This value represents the mass for the gold ball; now we determine the volume of the gold ball using its density.

Volume of gold ball = mass of gold ball/density of gold = 582.156 g/19.3 g/cm^3 = 30.1635 cm^3.

So this volume must correspond to a hollow sphere with an outer radius R = 3 cm and an unknown inner radius r.

Volume of the hollow ball can be represented as: 4/3 pi [R^3 - r^3].

Thus, 30.1635 cm^3 = 4/3 pi [3^3 - r^3].

30.1635 * 3/(4 * 3.14) = 27 - r^3.

Simplifying gives 7.2046 = 27 - r^3, resulting in r^3 = 19.7954.

Therefore, r = 2.7051 cm.

This indicates the thickness is the outer radius minus the inner radius: 3 - 2.7051 = 0.2949 cm.

Rounding to two significant figures yields

the thickness = 0.29 cm.

8 0
1 month ago
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