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Maslowich
3 years ago
10

What feature is forming ?

Chemistry
1 answer:
IRISSAK [1]3 years ago
3 0

Answer:

Earthquake

Explanation:

Notice how it is starting to shift allowing the earthquake to form.

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The mass of an unidentified metal sphere is 133 grams. Students determine the
Rainbow [258]

Given the data from the question, the identity of the unknown metal having a of mass 133 g is Cobalt

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

<h3>How to determine the density </h3>
  • Mass = 133 g
  • Volume of water = 25 mL
  • Volume of water + metal = 40 mL
  • Vol of metal = 40 – 25 = 15 mL
  • Density =?

Density = mass / volume

Density = 133 / 15

Density = 8.86 g/mL

Comparing the density of the unknown metal (i.e 8.86 g/mL) with those given in the chart in the question above, we can conclude that the unknown metal is Cobalt

Learn more about density:

brainly.com/question/952755

4 0
2 years ago
The methylene chloride solution is filtered through sodium sulfate to:
Vlad [161]

Answer:

C. Dry the methylene chloride by removing water

Explanation:

Anhydrous sodium sulfate is known for its high capacity to absorb water, for this reason it is widely used in laboratories as a drying agent.

Sodium sulfate is a neutral molecule so it cannot be used to neutralize and is very stable, so it is difficult to precipitate organic molecules.

5 0
3 years ago
Write empirical formula
andrew11 [14]

Answer:

Pb(ClO_{3})_{4}\\Pb(MnO_{4})_{4}\\Fe(ClO_{3})_{3}\\\Fe(MnO_{4})_{3}\\

Explanation:

Pb^{4+}(ClO_{3}^{-})_{4}--->Pb(ClO_{3})_{4}\\Pb^{4+}(MnO_{4}^{-})_{4}--->Pb(MnO_{4})_{4}\\Fe^{3+}(ClO_{3}^{-})_{3}--->Fe(ClO_{3})_{3}\\\Fe^{3+}(MnO_{4}^{-})_{3}--->Fe(MnO_{4})_{3}\\

3 0
3 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
How are simple cations and anions named ?
algol13
Simple cations are formally called by their element names with a suffixed Roman numeral in parentheses to indicate its charge. A simple anion has a name that is the original elemental name with the final syllable changed to -ide.
3 0
3 years ago
Read 2 more answers
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