Question

1 w(Ва)=80,1%, w(Н)=1,2%, w (O)=18,7% 2 w(Cr)=21,8%, w(N)=17,6%, w (O)=60,5% 3 w(Na)=39,3%, w(CI)=60,7%. 4 w(Са) =40%, w(С)=12%, w (O)=48,% 5 w(Fe)=48,3%, w(C)=10,3%, w (O 6 w(Co)=32,2%, w(N)=15,3%, w (O)=52,5% Решите задачи,найти отношение

Answer 1

Ответ:

Ba(OH)₂

Cr(NO₃)₃

NaCl

CaCO₃

?FeCO₃? - не хватает условий

Co(NO₃)₂

Объяснение:

1 Ba(OH)₂

1. Найдём количественные доли (массовые разделим на относительную атомную массу элементов)

80,1/137=0,5847

1,2/1=1,2

18,7/16=1,169

2. Разделим на минимальное число

0,5847/0,5847=1

1,2/0,5847=2

1,169/0,5847=2

3. Получаем BaH2O2 или Ba(OH)₂

2-6 Все аналогично

Answer 2

Task 1: Calculate the ratio of w(Ва), w(Н), and w(O)

To calculate the ratio of w(Ва), w(Н), and w(O), we need to add up their respective percentages and divide each individual percentage by the sum.

Given: - w(Ва) = 80.1% - w(Н) = 1.2% - w(O) = 18.7%

To find the ratio, we first calculate the sum of the percentages: 80.1% + 1.2% + 18.7% = 100%

Next, we divide each individual percentage by the sum: - w(Ва) ratio = w(Ва) / sum = 80.1% / 100% = 0.801 - w(Н) ratio = w(Н) / sum = 1.2% / 100% = 0.012 - w(O) ratio = w(O) / sum = 18.7% / 100% = 0.187

Therefore, the ratio of w(Ва), w(Н), and w(O) is approximately: - w(Ва) ratio = 0.801 - w(Н) ratio = 0.012 - w(O) ratio = 0.187

Task 2: Calculate the ratio of w(Cr), w(N), and w(O)

To calculate the ratio of w(Cr), w(N), and w(O), we follow the same process as in Task 1.

Given: - w(Cr) = 21.8% - w(N) = 17.6% - w(O) = 60.5%

Sum of the percentages: 21.8% + 17.6% + 60.5% = 100%

Calculating the ratios: - w(Cr) ratio = w(Cr) / sum = 21.8% / 100% = 0.218 - w(N) ratio = w(N) / sum = 17.6% / 100% = 0.176 - w(O) ratio = w(O) / sum = 60.5% / 100% = 0.605

Therefore, the ratio of w(Cr), w(N), and w(O) is approximately: - w(Cr) ratio = 0.218 - w(N) ratio = 0.176 - w(O) ratio = 0.605

Task 3: Calculate the ratio of w(Na) and w(CI)

Given: - w(Na) = 39.3% - w(CI) = 60.7%

Sum of the percentages: 39.3% + 60.7% = 100%

Calculating the ratios: - w(Na) ratio = w(Na) / sum = 39.3% / 100% = 0.393 - w(CI) ratio = w(CI) / sum = 60.7% / 100% = 0.607

Therefore, the ratio of w(Na) and w(CI) is approximately: - w(Na) ratio = 0.393 - w(CI) ratio = 0.607

Task 4: Calculate the ratio of w(Са), w(С), and w(O)

Given: - w(Са) = 40% - w(С) = 12% - w(O) = 48%

Sum of the percentages: 40% + 12% + 48% = 100%

Calculating the ratios: - w(Са) ratio = w(Са) / sum = 40% / 100% = 0.4 - w(С) ratio = w(С) / sum = 12% / 100% = 0.12 - w(O) ratio = w(O) / sum = 48% / 100% = 0.48

Therefore, the ratio of w(Са), w(С), and w(O) is approximately: - w(Са) ratio = 0.4 - w(С) ratio = 0.12 - w(O) ratio = 0.48

Task 5: Calculate the ratio of w(Fe), w(C), and w(O)

Given: - w(Fe) = 48.3% - w(C) = 10.3% - w(O) = 41.4%

Sum of the percentages: 48.3% + 10.3% + 41.4% = 100%

Calculating the ratios: - w(Fe) ratio = w(Fe) / sum = 48.3% / 100% = 0.483 - w(C) ratio = w(C) / sum = 10.3% / 100% = 0.103 - w(O) ratio = w(O) / sum = 41.4% / 100% = 0.414

Therefore, the ratio of w(Fe), w(C), and w(O) is approximately: - w(Fe) ratio = 0.483 - w(C) ratio = 0.103 - w(O) ratio = 0.414

Task 6: Calculate the ratio of w(Co), w(N), and w(O)

Given: - w(Co) = 32.2% - w(N) = 15.3% - w(O) = 52.5%

Sum of the percentages: 32.2% + 15.3% + 52.5% = 100%

Calculating the ratios: - w(Co) ratio = w(Co) / sum = 32.2% / 100% = 0.322 - w(N) ratio = w(N) / sum = 15.3% / 100% = 0.153 - w(O) ratio = w(O) / sum = 52.5% / 100% = 0.525

Therefore, the ratio of w(Co), w(N), and w(O) is approximately: - w(Co) ratio = 0.322 - w(N) ratio = 0.153 - w(O) ratio = 0.525

Please note that the calculations are based on the given percentages and the assumption that the sum of the percentages for each task is 100%.