Answer: The new volume of the air in the syringe is 7.3 mL.
Explanation:
Given: 
= 2.0 mL, 
= 1.02 atm, 
= ?, 
= 1.27 atm, 
Formula used to calculate the new volume in syringe is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the new volume of the air in the syringe is 7.3 mL.
I believe it is D my good friend.
this question is in reference to the formation and naming of ionic compounds. Specifically, they want you to give examples of three ionic compounds with a metal to nonmetal ratio of 2 to 1. That means we need to have two metal atoms to metal ions, which are typically cat ions for every one non metal atom or an ion. In order for this to occur, we need to have the metal with half the charge of the nonmetal or the non metal with double the charge of the metal. So an example might be something like sodium sulfide. Sodium has one valence electron. It can give up sulfur needs to valence electrons in order to achieve an octet. So we need to. Sodium seems to give up one electron each to total so that sulfide can achieve an octet. Another one might be potassium oxide. Similar scenario. We've got potassium giving up one valence electron oxygen requiring too. So we need to potassium to supply the to valence electrons that oxygen needs to achieve an octet and lithium. Also in Group one A and alkali metal wants to give up just one valence electron to achieve an octet well to achieve, I guess a duet to be more like helium, and so it gives up one. If we have two of them, then we can provide the to valence electrons that sulfur needs. So this is sodium sulfide, potassium oxide and lithium sulfide. Remember when we name Ionic compounds? We named the Cat Ion with the name of the element and the anti on with the name of the Element, but with the ending oven of ID, a suffix of ID because each one of the cat ions donated their valence electrons to the anti on so the an ion could achieve an octet. Then all of the's will have an octet of valence electrons. Sulfur had six sodium had one. There were two of them, so we have a total of eight.
Answer:
33.30 grams of CaCl2 will be required
Explanation:
Given,
Volume of solution, V= 250 ml
Molarity of solution, M= 1.20 mol/L
Molecular mass of CaCL2, S= 40+(35.5 X 2)= 111
We know,
Required mass, W= SVM/1000
Now,
W = (111 X 250 X 1.20)/1000
= 33300/1000
= 33.30
Therefore, 33.30 grams of Calcium Chloride will be required.
1. Aqeuous or dissolved
2.AgNO3+ NaCl are reactants
3. NaNO3 is precipitate
4. Reaction is endothermic(Delta stands for Heat)
5. Neither a reactant nor a product, MnO2 is a catalyst
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