Well the solvent is the liquid in a solution so your answer would be Solute, D. That is the one that would represent the sugar crystals being evenly mixed into a solution.

6 0

3 years ago

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Rn-222 has a half-life of 3.82 days. If 25.0 g of Radon-22 was originally present, approximately how many grams would be left af

horsena [70]

Answer:

Approximately 0.39 g or 0.4 g if you're rounding up

Explanation:

15/3.82 = 3.92

Let's round that up to 4

That means 15 days is around 4 half lives

4 half lives means 1/16 of the original mass will be left

25/16 = 0.390625

6 0

4 years ago

What is the % composition of Carbon in Chromium (iii) Carbonate

photoshop1234 [79]

Step 1 - Discovering the ionic formula of Chromium (III) Carbonate

Chromium (III) Carbonate is formed by the ionic bonding between Chromium (III) (Cr(3+)) and Carbonate (CO3(2-)):

$Cr^{3+}+CO^{2-}_3\rightarrow Cr_2(CO_3)_3$

Step 2 - Finding the molar mass of the substance

To find the molar mass, we need to multiply the molar mass of each element by the number of times it appears in the formula of the substance and, finally, sum it all up.

The molar masses are 12 g/mol for C; 16 g/mol for O and 52 g/mol for Cr. We have thus:

$\begin{gathered} C\rightarrow3\times12=36 \\ \\ O\rightarrow9\times16=144 \\ \\ Cr\rightarrow2\times52=104 \end{gathered}$

The molar mass will be thus:

$M=36+104+144=284\text{ g/mol}$

Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

$\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\ \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}$

The percent composition of Carbon is thus 12.7 %.

8 0

1 year ago