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LekaFEV
1 month ago
10

During the lab, you will have access to a range of acids and bases as well as universal pH indicator paper. Think about how you

can use these tools to develop a color chart for the red cabbage indicator. Then, write a hypothesis using the “if/then” format.
Hypothesis:
Chemistry
2 answers:
alisha [2.8K]1 month ago
6 0

Answer:

Explanation:bxbxbd

Alekssandra [2.8K]1 month ago
6 0

Answer:

This is a sample response meant for educational purposes. Refrain from copying it verbatim; doing so could hinder your understanding and lead to plagiarism issues.

Explanation:

By measuring the pH levels of a variety of acids and bases with universal pH testing paper, you’ll be able to use these readings to calibrate the pH indicator derived from cabbage.

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Determine how many grams of silver would be produced, if 12.83 x 10^23 atoms of copper react with an excess of silver nitrate. G
Anarel [2728]
1) The chemical equation is

Cu + 2AgNO3 ---> Cu (NO3)2 + 2Ag

2) Molar ratios are as follows:

1 mol Cu: 2 moles AgNO3: 1 mol Cu (NO3)2: 2 mol Ag

3) Converting 12.83 * 10^23 atoms of Cu to moles gives:

12.83 * 10^23 atoms / (6.02 * 10^23 atoms / mol) = 2.131 mol Cu

4) Using the ratios:

2.131 mol Cu * 2 mol Ag / 1 mol Cu = 4.262 mol Ag

5) To convert 4.262 mol of silver to grams, use the atomic weight of silver:

mass = moles × atomic mass = 4.262 mol * 107.9 g / mol = 459.9 grams

Answer: 459.9 g
5 0
29 days ago
0.475 g H, 7.557 gS, 15.107 g O. Express your answer as a chemical formula.
lorasvet [2668]

Response:

H₂SO₄

Clarification:

Given a compound consisting of 0.475 g H, 7.557 g S, and 15.107 g O, we must compute the empirical formula by following specific steps.

Step 1: Compute the total mass of the compound

Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g

Total mass = 23.139 g

Step 2: Determine the percentage composition.

H: (0.475g/23.139g) × 100% = 2.05%

S: (7.557g/23.139g) × 100% = 32.66%

O: (15.107g/23.139g) × 100% = 65.29%

Step 3: Divide each percentage by the element's atomic mass

H: 2.05/1.01 = 2.03

S: 32.66/32.07 = 1.018

O: 65.29/16.00 = 4.081

Step 4: Normalize all values by the smallest one

H: 2.03/1.018 ≈ 2

S: 1.018/1.018 = 1

O: 4.081/1.018 ≈ 4

Thus, the empirical formula for the compound is H₂SO₄.

7 0
10 days ago
What is another name for the molecular orbital theory of bonding in metals?
Alekssandra [2891]
The correct option is A. The band theory of metal explains how metals carry electricity by utilizing the electrons in their outer shells. When atomic orbitals of metals with similar energy levels merge, they create molecular orbitals and form bands. These bands facilitate the movement of electrons within metals, enabling them to conduct electricity.
4 0
12 days ago
A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr
KiRa [2853]

Answer:

Explanation:

Given data:

Initial temperature T₁ = 25.2°C = 298.2K

Initial pressure P₁ = 0.6atm

Final temperature = 72.4°C = 345.4K

What we need to find:

Final pressure = ?

To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:

\frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.

Now we can substitute the known variables:

\frac{0.6}{298.2}   = \frac{P_{2} }{345.4}

P₂ = 0.7atm

3 0
1 month ago
Which statement is TRUE regarding the macroscopic and
lions [2782]

Answer:

Chemists observe phenomena on a macroscopic level which informs their understanding of microscopic aspects.

Explanation:

Many critical chemical insights arise from macroscopic observations because most scientific instruments currently cannot directly evidence microscopic events. Data gathered from these larger-scale observations can yield valuable insights into the nature of specific microscopic interactions.

This is particularly true in atomic structure studies. The majority of evidence that contributed to our understanding of atomic structure was obtained from macroscopic observations and subsequently provided crucial information regarding the atom's microscopic configuration.

7 0
1 month ago
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