Express the equation x^2 = 4y in polar coordinates.
Step 1: Remind yourself of the formula x = rcosθ and y = rsinθ. Step 2: Rewrite the given equation and replace x by rcosθ and y by rsinθ x^2 = 4y (rcosθ)^2 = 4(rsin θ) Step 3: Calculate r r^2 (cosθ)^2 = 4rsinθ r = 4 sinθ (cosθ)^2 = 4 secθ tanθ
Find the empirical formula for the oxide that contains 42.05 g of nitrogen and 95.95 g of oxygen.
It is really important to convert all the given amount into moles. The ratio of the moles of each element will provide the ratio of the atoms of each element. Step 1: Convert the mass of each element to moles of each element using the atomic masses. (42.05 g N) (1 mol/ 14.01 g N) = 3.001 mol N (95.95 g O) (1 mol/ 16.00 g O) = 5.997 mol O Step 2: Find the ratio or the moles of each element by dividing the number of moles of each by the smallest number of moles. The moles of nitrogen is the fewest. So we will divide each of the mole above by 3.001 (3.001 mol N/ 3.001) = 1 mol N (5.997 mol O/ 3.001) = 1.998 mol O Step 3: Use the mole ratio to write the empirical formula. Since there cannot be partial atoms in the empirical formula, the mole ratios must be whole numbers. 1.998 is sufficiently close to 2 that we can round. Thus the empirical formula is: NO2
Solve the equation 5(3x + 6) - (x + 3) = -4(4x + 5) + 13
Step 1: Distribution 15x + 30 - x - 3 = -16x -20 +13 Step 2: Grouping x-term and non x-term (15x - x) + (30-3) = -16x + (13-20) -> 14x + 27 = -16x - 7 Step 3: Eliminate x-term on one side (I choose the right side) and eliminate the constant number on the other side( so it has to be the right side). Add 16x on both side to eliminate x-term on the right side. Add -27 on both side to eliminate the constant number on the left side. 14x + 27 + (16x - 27) = -16x -7 + (16x - 27) Step 4: Calculate all the term from each side 30x = -34 Step 5: Find x by dividing both side to 30 30x/30 = -34/30 -> x = -34/30 -> simplification x = -17/15