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Kryger [21]
3 years ago
7

During an experiment, it is observed that iron powder sinks in water and sulfur powder floats in water. What will most likely be

observed when sulfur powder and iron powder are mixed together and then added to water?
Sulfur will float and iron will sink because each will retain its properties.
Iron will float and sulfur will sink because each will lose its properties.
Both will float because they are physically mixed.
Both will float because they react chemically.

someone pls help me, i literally don't understand chem
Ill give 50 points
Chemistry
2 answers:
lutik1710 [3]3 years ago
7 0

Sulfur will float, and iron will sink because each will retain its properties.

“Iron will float, and sulfur will sink because each will lose its properties” is <em>incorrect</em>. A substance will lose its properties only if it reacts to form a new substance.

“Both will float because they are physically mixed” is <em>incorrect</em>. The substances in a mixture retain their properties.

“Both will float because they react chemically” is <em>incorrect</em>. Iron and sulfur do not react at room temperature. Even if they did, the iron sulfide would sink.

irga5000 [103]3 years ago
6 0

Answer:

Sulfur will float and iron will sink because each will retain its properties.

Explanation:

This is the correct answer because sulfur powder and iron powder are two different things, with different properties. Even if they are added to water and mixed around, they still retain their properties since no actual chemical activity was performed on them (they were just mixed by physical means). Hence, the sulfur will float back up, as it does initially, and the iron will sink, as it had initially.

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A cyanide solution with avolume of 12.73 mL was treated with 25.00 mL of Ni2+solution (containing
natima [27]

Answer:

The molarity of this solution is 0,09254M

Explanation:

The concentration of the Ni²⁺ solution is:

Ni²⁺ + EDTA⁴⁻ → Ni(EDTA)²⁻

0,03935L × 0,01307M = 5,143x10⁻⁴ moles Ni²⁺ ÷ 0,03010L =<em>0,01709M Ni²⁺</em>

25,00 mL of this solution contain:

0,01709M × 0,02500L = 4,2716x10⁻⁴ moles of Ni²⁺

The moles of Ni²⁺ that are in excess and react with EDTA⁴⁻ are:

0,01015L × 0,01307M = 1,3266x10⁻⁴ moles of Ni²⁺

Thus, moles of Ni²⁺ that react with CN⁻ are:

4,2716x10⁻⁴ - 1,3266x10⁻⁴ = 2,9450x10⁻⁴ moles of Ni²⁺

For the reaction:

4CN⁻ + Ni²⁺ → Ni(CN)₄²⁻

Four moles of CN⁻ react with 1 mole of Ni²⁺:

2,9450x10⁻⁴ moles of Ni²⁺ × \frac{4 mol CN^-}{1 molNi^{2+}} = <em>1,178x10⁻³ moles of CN⁻</em>

As the volume of cyanide solution is 12,73mL. The molarity of this solution is:

<em>1,178x10⁻³ moles of CN⁻ ÷ 0,01273L = </em><em>0,09254M</em>

I hope it helps!

5 0
2 years ago
4
matrenka [14]

Answer:

Yes.

Explanation:

Yes, this difference of readings will definitely affect the results of the experiment as well as the E values because the readings taken by both students are different from one another. There is a fault in one of the thermometer because both shows different readings of temperature of the same solution. This will affect the overall experiment and due to this error, we are unable to tell that which one reading is correct so the answer is uncertain or unsure.

5 0
2 years ago
You will need to prepare 12 mL of 25% Sodium Phosphate Buffer (pH 4) solution for Activity 2. What volume of the stock Sodium Ph
zhuklara [117]

Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>

The process of preparing solutions from stock solutions of higher concentration is known as dilution.

Dilution is done with the aid of the dilution formula given below:

  • C1V1 = C2V2

where

  • C1 is the concentration of stock solution
  • V1 is the volume of stock solution required to prepare a diluted solution
  • C2 is the concentration of the diluted solution prepared
  • V2 is the final volume of the diluted solution

From the data provided:

C1 is not given

V1 is unknown

C2 = 25%

V2 = 12 mL

  • Assuming C1 is 50% solution

Volume of stock, V1, required is calculated as follows:

V1 = C2V2/C1

V1 = 25 × 12 /50

V1 = 6 mL

Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

Learn more about dilution formula at: brainly.com/question/7208546

6 0
2 years ago
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murzikaleks [220]
The answer is A) Aluminum Bromide 
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3 0
2 years ago
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Rainbow [258]

Answer:

c

Explanation:

use %composition given to calculate emperical 4mular

8 0
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