This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here's the solution to question 11: Let O be the center of the rectangular base PQRS. Since the pyramid is a right pyramid, the vertex V is directly above O, meaning VO is the height of the pyramid and is perpendicular to the base PQRS. Given: VP = VQ = VR = VS = 18 cm PQ = 16 cm QR = 12 cm a) The length of the projection of line VP on the plane PQRS. Step 1: Identify the projection. The projection of line VP onto the plane PQRS is the line segment connecting P to the foot of the perpendicular from V to the plane, which is O. So, the projection is OP. Step 2: Calculate the length of the diagonal of the base. In the rectangular base PQRS, consider the right-angled triangle PQR. Using the Pythagorean theorem: PR^2 = PQ^2 + QR^2 PR^2 = 16^2 + 12^2 PR^2 = 256 + 144 PR^2 = 400 PR = sqrt(400) = 20 cm Step 3: Calculate the length of OP. Since O is the center of the rectangle, it is the midpoint of the diagonal PR. OP = (1)/(2) PR OP = (1)/(2) × 20 OP = 10 cm The length of the projection of line VP on the plane PQRS is 10 cm. b) The size of the angle between line VP and the plane PQRS. Step 1: Identify the angle. The angle between line VP and the plane PQRS is the angle between VP and its projection OP. This is VPO. Step 2: Use trigonometry in the right-angled triangle VOP. Triangle VOP is a right-angled triangle with the right angle at O. We know VP = 18 cm (hypotenuse) and OP = 10 cm (adjacent to VPO). ( VPO) = (OP)/(VP) ( VPO) = (10)/(18) ( VPO) = (5)/(9) VPO = ((5)/(9)) VPO ≈ 56.25^ The size of the angle between line VP and the plane PQRS is 56.3^ (to one decimal place). c) The size of the angle between the planes VQR and PQRS. Step 1: Identify the line of intersection and perpendiculars. The line of intersection of plane VQR and plane PQRS is the line segment QR. Let M be the midpoint of QR. Since triangle VQR is an isosceles triangle (VQ = VR = 18 cm), the line segment VM from the vertex V to the midpoint M of the base QR is perpendicular to QR. So, VM QR. In the rectangular base PQRS, the line segment OM (connecting the center O to the midpoint M of QR) is perpendicular to QR. So, OM QR. Therefore, the angle between the planes VQR and PQRS is VMO. Step 2: Calculate the length of OM. Since O is the center of the rectangle and M is the midpoint of QR, OM is parallel to PQ and its length is half of PQ. OM = (1)/(2) PQ OM = (1)/(2) × 16 OM = 8 cm Step 3: Calculate the height of the pyramid VO. In the right-angled triangle VOP (from part b), we can find VO using the Pythagorean theorem: VO^2 + OP^2 = VP^2 VO^2 + 10^2 = 18^2 VO^2 + 100 = 324 VO^2 = 324 - 100 VO^2 = 224 VO = sqrt(224) = sqrt(16 × 14) = 4sqrt(14) cm Step 4: Use trigonometry in the right-angled triangle VOM. Triangle VOM is a right-angled triangle with the right angle at O. We know VO = 4sqrt(14) cm (opposite to VMO) and OM = 8 cm (adjacent to VMO). ( VMO) = (VO)/(OM) ( VMO) = 4sqrt(14)8 ( VMO) = sqrt(14)2 VMO = (sqrt(14)2) VMO ≈ 61.86^ The size of the angle between the planes VQR and PQRS is 61.9^ (to one decimal place). What's next?