What is the volume in mL of a balloon filled with 32g of helium gas at STP?
To begin this problem, we must figure out which equation to use. The question states the balloon is at STP, which means standard temperature and pressure. This tells us we can use an equation with temperature (T) and pressure (P). We also know the equation should have volume (V), since that's what we're trying to find. Finally, we are given grams, a mass, which through calculations can be converted to moles (n). With all of these variables, we know we can use the Ideal Gas Law, PV=nRT. Let's start by converting 32g of Helium into moles. From the periodic table, we can see that one mole of He equals about 4 grams. Now we can simply divide the given 32g by 4g to get 8 moles of He. Next, we must understand what STP tells us. The standard temperature is 273K and the standard pressure is 1 atm. Finally, the value R is a constant that will always equal 0.0821. This leaves us with a basic algebra problem by plugging in the known information to PV=nRT... (1atm)(V)=(8 moles)(.0821)(273K). After solving for V, we get V = 179.3 L. The last step would be to convert liters into mL, by multiplying our answer by 1000. The final volume would be 179,306 mL, or since the problem only gave us two significant figure in the 32g, 180,000 mL rounded to the correct number of sig figs.
How many atoms are in 55g of magnesium chloride?
First we need to balance our compound. To do this, we'll look at the charges of a magnesium (Mg) cation and a chloride (Cl) anion. The periodic table shows that Mg is in the second column, meaning it has a +2 charge. Cl is in the 7/17th column, meaning it has a -1 charge (needs to gain one electron to reach its octet of 8 electrons). Now that we know their chargers, we can balance them. For every one magnesium ion (+2 charge), we need two chloride ions (-1 charge each). This leaves us with our balanced compound: MgCl2. Now we need to convert our mass (55g) into moles. To do this, we should figure out how many grams there are in one mole of sodium chloride. Looking at the periodic table, we will find that the atomic mass of sodium, or Mg, is 24.3, and the atomic mass of chloride, or Cl, is 35.45. Since our ionic charge ratio was 1:2, we need to account for that by adding in two 35.45's into our MgCl2 molar mass. Now we're left with 24.3+35.45+35.45, equaling 95.2. This tells us that for every one mole, there are 95.2 grams. Going back to the original problem, we can convert 55g of MgCl2 into moles of MgCl2 by dividing 55/95.2, telling us that 55g of MgCl2 equaled .57 moles MgCl2. The last step is to convert this into atoms, using Avogadro's number (6.02*10^23). We'll simply multiply .57 by this number to reach our final answer, 3.47*10^23. Since our original problem gave us only two significant figures, we should round up to 3.5*10^23 atoms.
Solve for x in the following equation: 13x - 3(6 + y) = 4 + 3x + y
To solve for x, we need to isolate all of the x's to one side of the equation, and everything else to the other side. To move something to the other side of the equation, we should either subtract it from both sides (if it is a negative number) or add to both sides (if it is positive). This will leave us with 13x - 3x = 4 + y + 3(6+y). To further simplify, we can combine like terms; that is, subtract the x's from each other to get 10x on the left side, and add the y's together. Since the (6+y) is enclosed in parenthesis, we need to multiply both 6 and y by the 3 that is located outside of the parenthesis, leaving us with 4 + y + 18 + 3y on the right side of the equation. By adding like terms, we will get a simplified 22 + 4y. Putting it all together, our whole equation now reads 10x = 22 + 4y. For our final answer, we should divide everything by 10 so we are left with just the letter x. X now equals (22+4y)/10, or even more simplified, (11+2y)/5.