Step 7- Communicate. Present/share your results. Replicate.
Step 1- Question.
Step 2-Research.
Step 3-Hypothesis.
Step 4-Experiment.
Step 5-Observations.
Step 6-Results/Conclusion.
Answer:
Pentacarbon heptasilicide.
Explanation:
In order to name the following compound, we need to identify whether it is molecular or ionic.
Molecular compounds consist of non-metal atoms, while ionic compounds would have metal cations in their composition.
In the given compound,
, we have two non-metals, carbon and silicon, meaning we should follow the molecular compound naming rules. The rules involve using prefixes to state the number of individual atoms.
The two prefixes required here are 'penta' for 'five' to indicate 5 carbon atoms present and 'hepta' for 'seven' to indicate 7 silicon atoms present.
The first part of the name would be pentacarbon (notice that the standard name for the first element is used). The second part would be heptasilicide (notice that the second atom would have an ending of -ide followed by the standard beginning of silicon).
The lighter components are able to rise higher in the column before they are cooled to their condensing temperature, allowing them to be removed at slightly higher levels.
I hope this helps
Answer:
Exam 3 Material
Homework Page Without Visible Answers
This page has all of the required homework for the material covered in the third exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.
Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.
These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.
Explanation:
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)