Question

Daнo m p-pa = 250 г. w pв - 10% mp-pa²= 500 г. w pв²=20%+m(H²0)=100г.w pв³=?

Answer 1

25+100+100=225гр третьего раствора

Answer 2

To solve this problem, let's break it down step by step.

Given:

1. m * p-ра = 250 г
2. w рв = 10% of m * p-ра² = 500 г
3. w рв² = 20% + m(H₂0) = 100 г

We need to find w рв³.

Step 1: Find the value of m * p-ра. m * p-ра = 250 г (given)

Step 2: Find the value of m. Since m * p-ра = 250 г and m * p-ра² = 500 г, we can find m by dividing the second equation by the first one. m = (m * p-ра²) / (m * p-рра) = 500 г / 250 г = 2

Step 3: Find the value of p-ра. p-ра = 250 г / m = 250 г / 2 = 125 г

Step 4: Find the value of w рв³. w рв³ = 100% - (w рв + w рв²) w рв³ = 100% - (10% of m * p-ра² + 20% + m(H₂0)) w рв³ = 100% - (10% * 500 г + 20% + 2 * 100 г) w рв³ = 100% - (50 г + 20 г + 200 г) w рв³ = 100% - 270 г w рв³ = 100 г

So, w рв³ = 100 г.