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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To multiply binary numbers, we use a process similar to long multiplication with decimal numbers. Here are the rules for binary multiplication: • 0 × 0 = 0 • 0 × 1 = 0 • 1 × 0 = 0 • 1 × 1 = 1 And for binary addition: • 0 + 0 = 0 • 0 + 1 = 1 • 1 + 0 = 1 • 1 + 1 = 10_2 (which means 0 with a carry of 1) • 1 + 1 + 1 = 11_2 (which means 1 with a carry of 1) We need to calculate 1111011_2 × 1101_2. Step 1: Write down the numbers for long multiplication. ccccccccl & & 1 & 1 & 1 & 1 & 0 & 1 & 1_2 \\ × & & & & & 1 & 1 & 0 & 1_2 \\ Step 2: Multiply the top number (1111011_2) by each digit of the bottom number (1101_2), starting from the rightmost digit. Remember to shift each new product to the left. Multiply by the rightmost 1: 1111011_2 × 1_2 = 1111011_2 Multiply by the next digit 0: 1111011_2 × 0_2 = 0000000_2. Shift this one place to the left: 00000000_2 Multiply by the next digit 1: 1111011_2 × 1_2 = 1111011_2. Shift this two places to the left: 111101100_2 Multiply by the leftmost digit 1: 1111011_2 × 1_2 = 1111011_2. Shift this three places to the left: 1111011000_2 Now, arrange these partial products for addition: cccccccccccc & & & & & 1 & 1 & 1 & 1 & 0 & 1 & 1_2 \\ & & & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0_2 \\ & & & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & 0_2 \\ + & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0_2 \\ Step 3: Add the partial products together, column by column, from right to left. Remember to carry over any 1s when the sum is 10_2 or 11_2. Let's add them carefully: ccccccccccccl Carries: & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & & \\ & & & & & 1 & 1 & 1 & 1 & 0 & 1 & 1_2 \\ & & & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0_2 \\ & & & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & 0_2 \\ + & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0_2 \\ & 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1_2 \\ Let's break down the addition column by column: • Rightmost column: 1+0+0+0 = 1. • Second column: 1+0+0+0 = 1. • Third column: 0+0+1+0 = 1. • Fourth column: 1+0+1+0 = 10_2. Write 0, carry 1. • Fifth column: 1+0+0+1 + (carry 1) = 1+0+0+1+1 = 11_2. Write 1, carry 1. • Sixth column: 1+0+1+1 + (carry 1) = 1+0+1+1+1 = 100_2. Write 0, carry 10_2 (which is 2_10). This is 1+1+1+1 = 100_2. Write 0, carry 10_2. Let's re-evaluate this step. It's easier to add two numbers at a time. Let's add the partial products step-by-step: First, add the first two partial products: cccccccccl & & & & 1 & 1 & 1 & 1 & 0 & 1 & 1_2 \\ + & & & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0_2 \\ & & & & 1 & 1 & 1 & 1 & 0 & 1 & 1_2 \\ Next, add the third partial product (111101100_2) to the result: cccccccccl Carries: & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & & \\ & & & & 1 & 1 & 1 & 1 & 0 & 1 & 1_2 \\ + & & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & 0_2 \\ & & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1_2 \\ Let's verify this addition: 1+0=1 1+0=1 0+1=1 1+1=0 (carry 1) 1+0+1 (carry)=0 (carry 1) 1+1+1 (carry)=1 (carry 1) 1+1 (carry)=0 (carry 1) 0+1 (carry)=1 0+0=0 0+0=0 Result: 100110011_2. (This is the sum of the first three lines) Finally, add the fourth partial product (1111011000_2) to this new sum: cccccccccl Carries: & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & & \\ & & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1_2 \\ + & 1 & 1 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 0_2 \\ 1 & 1 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1_2 \\ Let's verify this final addition: 1+0=1 1+0=1 0+0=0 0+1=1 1+1=0 (carry 1) 1+0+1 (carry)=0 (carry 1) 0+1+1 (carry)=0 (carry 1) 0+1+1 (carry)=0 (carry 1) 1+1 (carry)=0 (carry 1) 0+1 (carry)=1 0+0=0 Result: 100000100111_2. Let's re-do the final addition one more time, very carefully. Sum of first three: 100110011_2 Fourth partial product: 1111011000_2 Align them: ` 000100110011 (Sum of first three, padded) + 111101100000 (Fourth partial product, padded) -------------- ` Adding column by column from right to left: 1. 1+0 = 1 2. 1+0 = 1 3. 0+0 = 0 4. 0+1 = 1 5. 1+1 = 0 (carry 1) 6. 0+0+1 (carry) = 1 7. 0+1 = 1 8. 1+1 = 0 (carry 1) 9. 0+1+1 (carry) = 0 (carry 1) 10. 0+1+1 (carry) = 0 (carry 1) 11. 0+1 (carry) = 1 The final result is 100001101011_2. Let's double check with decimal conversion: 1111011_2 = 1 · 2^6 + 1 · 2^5 + 1 · 2^4 + 1 · 2^3 + 0 · 2^2 + 1 · 2^1 + 1 · 2^0 = 64+32+16+8+0+2+1 = 123_10 1101_2 = 1 · 2^3 + 1 · 2^2 + 0 · 2^1 + 1 · 2^0 = 8+4+0+1 = 13_10 123_10 × 13_10 = 1599_10 Now convert 1599_10 to binary: 1599 ÷ 2 = 799 R 1 799 ÷ 2 = 399 R 1 399 ÷ 2 = 199 R 1 199 ÷ 2 = 99 R 1 99 ÷ 2 = 49 R 1 49 ÷ 2 = 24 R 1 24 ÷ 2 = 12 R 0 12 ÷ 2 = 6 R 0 6 ÷ 2 = 3 R 0 3 ÷ 2 = 1 R 1 1 ÷ 2 = 0 R 1 Reading the remainders from bottom up: 11000111111_2. My binary multiplication result 100001101011_2 is still not matching 11000111111_2. There is a consistent error in my binary addition. Let's re-do the addition of the four lines using a different method, or just be extremely careful. Let's write the four lines again: Line 1: 0000001111011 Line 2: 0000000000000 (shifted) Line 3: 0000111101100 (shifted) Line 4: 0111101100000 (shifted) Let's add them column by column, keeping track of carries. ` Carries: 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 1 1 (Line 1) 0 0 0 0 0 0 0 0 0 0 0 0 0 (Line 2) 0 0 0 0 1 1 1 1 0 1 1 0 0 (Line 3) + 0 1 1 1 1 0 1 1 0 0 0 0 0 (Line 4) --------------------------- 1 0 0 1 1 0 0 1 1 1 1 1 1 ` Let's trace the carries and sums: 1. Rightmost column: 1+0+0+0 = 1. Sum: 1. Carry: 0. 2. Next column: 1+0+0+0 = 1. Sum: 1. Carry: 0. 3. Next column: 0+0+1+0 = 1. Sum: 1. Carry: 0. 4. Next column: 1+0+1+0 = 10_2. Sum: 0. Carry: 1. 5. Next column: 1+0+0+1 + (carry 1) = 1+0+0+1+1 = 11_2. Sum: 1. Carry: 1. 6. Next column: 1+0+1+1 + (carry 1) = 1+0+1+1+1 = 100_2. Sum: 0. Carry: 10_2 (which is 2_10). This is where the "basic one" explanation gets tricky. 1+1+1+1 = 100_2. So, write 0, carry 10_2. Let's write the carries as 1s and 0s. 1+1+1+1 = 100_2. So, 0 in the current column, and 10_2 (which is 2_10) is carried over. This means two 1s are carried to the next column. Let's use a simpler method for addition, adding two numbers at a time. 1. 1111011_2 2. 00000000_2 3. 111101100_2 4. 1111011000_2 Sum of (1) and (2): 1111011_2 Sum of (1111011_2) and (3) (111101100_2): ` 0001111011 + 00111101100 -------------- 01011001111 (This was the result I got earlier, 687_10, which was wrong) ` Let's re-do this specific addition: 0001111011 00111101100 ------------- 1+0 = 1 1+0 = 1 0+1 = 1 1+1 = 0 (carry 1) 1+0+1 (carry) = 0 (carry 1) 1+1+1 (carry) = 1 (carry 1) 1+1 (carry) = 0 (carry 1) 0+1 (carry) = 1 0+0 = 0 0+0 = 0 Result: 01010010111_2. (This is 663_10, still not 615_10) The error is in my binary addition. Let's use the standard long addition format with carries written above. Adding 1111011_2 and 111101100_2: ` Carries: ✂️ _That answer was long and got cut off. Reply continue and I'll finish it._