**Water Quality Analysis of the Thames River in London**The following information presents the results of a water quality analysis performed on the Thames River in London. The analysis considers water treated with 60 mg/L of alum to remove turbidity. For this study, ignore side reactions with phosphorus and assume that all the alkalinity is due to \( \text{HCO}_3^- \).### Objectives:Calculate the following:a. Consumption of \( \text{HCO}_3^- \) in mmol/Lb. Alkalinity consumed (mg/L as \( \text{CaCO}_3 \))c. Remaining alkalinity (mg/L as \( \text{CaCO}_3 \))### Analysis Data:| Constituent | Expressed as | mg/L ||-----------------------|--------------|------|| Total iron | Fe | 1.8 || Copper | Cu | 0.05 || Chromium | Cr | 0.01 || Total alkalinity | \( \text{CaCO}_3 \) | 130 || Chloride | Cl | 52 || Total Phosphate | \( \text{PO}_4 \) | 1 || Silica | \( \text{SiO}_2 \) | 14 || Suspended Solids | | 43 || Total Solids | | 495 || pH | | 7.4 |### Calculated Solutions:**Answers:**- a: 1.05 mmol/L \( \text{HCO}_3^- \)- b: 52.6 mg/L as \( \text{CaCO}_3 \)- c: 77.4 mg/L as \( \text{CaCO}_3 \)### Problem-Solving Process:To solve for these values:1. **Convert total alkalinity (as \( \text{CaCO}_3 \)) to \( \text{HCO}_3^- \)**: - Alkalinity in mg/L as \( \text{CaCO}_3 \) can be converted to mmol/L as \( \text{HCO}_3^- \) using the equivalent weight. - The equivalent weight for \( \text{CaCO}_3 \) is approximately 50 mg/meq. 2.

Answered: Image /qna-images/answer/58009df5-7587-4928-8c03-4a5a3f3c3d63.jpg

Shown below are the results of water quality analyses of the Thames River in London. The water is treated with 60 mg/L of alum to remove turbidity. Ignore side reactions with phosphorous and assume all the alkalinity is HCO3. a. HCO3 consumed in mmol/L b. Alkalinity consumed in mg/L as CaCO3 c. Alkalinity remaining in mg/L as CaCO3 Answers: a: 1.05 mmol/L HCO3- b: 52.6 mg/L as CaCO3 c: 77.4 mg/L as CaCO3 Constituent Total iron Copper Chromium Total alkalinity Chloride Total Phosphate Silica Suspended Solids Total Solids pH Expressed as Fe Cu Cr CaCO Cl PO4 SiO₂ mg/L 1.8 0.05 0.01 130 52 1 14 43 495 7.4 I have the answers but not sure how they were found so please work problem to get these answers.

Shown below are the results of water quality analyses of the Thames River in London. The water is treated with 60 mg/L of alum to remove turbidity. Ignore side reactions with phosphorous and assume all the alkalinity is HCO3. a. HCO3 consumed in mmol/L b. Alkalinity consumed in mg/L as CaCO3 c. Alkalinity remaining in mg/L as CaCO3 Answers: a: 1.05 mmol/L HCO3- b: 52.6 mg/L as CaCO3 c: 77.4 mg/L as CaCO3 Constituent Total iron Copper Chromium Total alkalinity Chloride Total Phosphate Silica Suspended Solids Total Solids pH Expressed as Fe Cu Cr CaCO Cl PO4 SiO₂ mg/L 1.8 0.05 0.01 130 52 1 14 43 495 7.4 I have the answers but not sure how they were found so please work problem to get these answers.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

**Water Quality Analysis of the Thames River in London**

The following information presents the results of a water quality analysis performed on the Thames River in London. The analysis considers water treated with 60 mg/L of alum to remove turbidity. For this study, ignore side reactions with phosphorus and assume that all the alkalinity is due to \( \text{HCO}_3^- \).

### Objectives:
Calculate the following:
a. Consumption of \( \text{HCO}_3^- \) in mmol/L
b. Alkalinity consumed (mg/L as \( \text{CaCO}_3 \))
c. Remaining alkalinity (mg/L as \( \text{CaCO}_3 \))

### Analysis Data:

| Constituent | Expressed as | mg/L |
|-----------------------|--------------|------|
| Total iron | Fe | 1.8 |
| Copper | Cu | 0.05 |
| Chromium | Cr | 0.01 |
| Total alkalinity | \( \text{CaCO}_3 \) | 130 |
| Chloride | Cl | 52 |
| Total Phosphate | \( \text{PO}_4 \) | 1 |
| Silica | \( \text{SiO}_2 \) | 14 |
| Suspended Solids | | 43 |
| Total Solids | | 495 |
| pH | | 7.4 |

### Calculated Solutions:

**Answers:**
- a: 1.05 mmol/L \( \text{HCO}_3^- \)
- b: 52.6 mg/L as \( \text{CaCO}_3 \)
- c: 77.4 mg/L as \( \text{CaCO}_3 \)

### Problem-Solving Process:

To solve for these values:
1. **Convert total alkalinity (as \( \text{CaCO}_3 \)) to \( \text{HCO}_3^- \)**:
- Alkalinity in mg/L as \( \text{CaCO}_3 \) can be converted to mmol/L as \( \text{HCO}_3^- \) using the equivalent weight.
- The equivalent weight for \( \text{CaCO}_3 \) is approximately 50 mg/meq.

2.

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