Question

W(ca) =40пр. W(c) =12пр.W(o)=48пр.СахСуОz=? ​

Answer 1

Ответ:

CaCO3

Объяснение:

Используем метод подбора, одно из возможных соединений CaCO3

w(Ca) = M(Ca) / M(CaCO3) = 40/100 *100% = 40(%)

w(C) = M(C) / M(CaCO3) = 12/100 * 100% = 12 (%)

w(O) = 100% - w(C) - w(Ca) = 100 - 52 = 48 (%))

Данное вещество подходит.

Answer 2

It seems that you have provided some abbreviations that are not clear to me. However, based on the context, it appears that you are referring to some substances or compounds and their respective masses.

I will try to interpret the abbreviations and solve the equation for you. Please confirm if my interpretation is correct:

• W(ca) = 40 пр: This could be the mass (in пр, which I assume stands for "parts" or "units") of a substance or compound labeled as "ca."
• W(c) = 12 пр: This could be the mass of another substance or compound labeled as "c."
• W(o) = 48 пр: This could be the mass of a third substance or compound labeled as "o."
• СахСуОz = ? : This could be the label for a compound whose mass is unknown and needs to be determined.

Assuming my interpretation is accurate, to find the mass of СахСуОz, we can sum up the masses of all the substances involved:

СахСуОz = W(ca) + W(c) + W(o) = 40 пр + 12 пр + 48 пр = 100 пр

So, the mass of СахСуОz is 100 пр. However, I can't determine the actual identity of the compound СахСуОz since the abbreviations provided are not commonly used or clear. If you can provide more information or clarification, I would be glad to assist further.