From 22ec9a6b0edfb5afda91fdf400d86d3bd75e29b9 Mon Sep 17 00:00:00 2001 From: Dane Sabo Date: Tue, 3 Sep 2024 14:17:33 -0400 Subject: [PATCH] vault backup: 2024-09-03 14:17:32 --- .../2024-09-03 Homework 1.md | 4 +++- 1 file changed, 3 insertions(+), 1 deletion(-) diff --git a/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md b/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md index 8c8116b84..f08e07319 100644 --- a/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md +++ b/300s School/302. NUCE 2100 - Fundamentals of Nuclear Engineering/2024-09-03 Homework 1.md @@ -30,8 +30,10 @@ $\frac{ ^{59}\text{Co}}{^{12}\text{C}} = \frac{58.93319}{12.00000} = 4.91110 \te $10 \text{g} \times \frac{1 \text{ mol } ^{12}\text{C}}{12 \text{g}} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol } ^{12}\text{C}} = 5.0184 \times 10^{23} \text{ atoms of } ^{12} \text{C}$ **4. A beaker contains 50 g of ordinary water.** *a. How many moles of water are present?* - $10 \text{g} \times \frac{1 \text{ mol } ^{12}\text{C}}{12 \text{g}} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol } ^{12}\text{C}} = 5.0184 \times 10^{23} \text{ atoms of } ^{12} \text{C}$ + $50 \text{g} \times \frac{1 \text{ mol } H_2O}{18.01528 \text{g}} = 2.77542 \text{ moles of water}$ *b. How many hydrogen atoms?* + $2.77542 \text{ moles of water } \times \frac{2 \text{mol} H}{1 \text{mol} H_2O} \times \frac{0.6022045 \times 10^{24} \text{ atoms}}{1 \text{ mol }}$ + *c. How many deuterium atoms?* **5. Find the mass of an atom of $^{235}\text{U}$** *a. in amu;*