How do you describe how sugar dissolves in tea using the big idea of particles?

2 answers:

4 0

If you use big pieces of sugar, the particles will dissolve harder...As the pieces are smaller, as much they dissolve faster.

Komok [63]3 years ago

3 0

Sugar is an ionic compound and tea is most likely a polar substance. When the sugar is placed in the tea the molecules of the tea attack the sugar molecules using their negative and positive pulls. The negative sides of the tea molecules pull at the positive parts of the sugar, and vice versa. <span />

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1.65g of zinc is used to make 8g of zinc iodide. How much iodine is required for this reaction?

creativ13 [48]

Answer:

6.45 g of iodine, I₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + I₂ —> ZnI₂

Next, we shall determine the mass of Zn and I₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of Zn = 65 g/mol

Mass of Zn from the balanced equation = 1 × 65 = 65 g

Molar mass of I₂ = 127 × 2 = 254 g/mol

Mass of I₂ from the balanced equation = 1 × 254 = 254 g

SUMMARY:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Finally, we shall determine the mass of f I₂ needed to react with 1.65 g of Zn. This can be obtained as follow:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Therefore, 1.65 g of Zn will react with = (1.65 × 254)/65 = 6.45 g of I₂.

Thus, 6.45 g of iodine, I₂ is needed for the reaction.

4 0

3 years ago

Unrestricted populations of organisms experience _____.

Fudgin [204]

Unrestricted populations of organisms experience Exponential growth.

As a population reaches is carrying capacity there is an increase in competition for All of the above.
Hope I helped.

5 0

3 years ago

Read 2 more answers

What percentage of children who are raised in abusive households grow up to be abusive parents themselves?

Gwar [14]

33%, according to Quizlet

8 0

3 years ago

Write the balanced nuclear equation for β− decay of silicon−32. Include both the mass numbers and the atomic numbers with each n

weqwewe [10]

Answer:

$^{32}_{14}Si\rightarrow ^{32}_{15}P+e^-+\bar{v}_e$

Explanation:

Beta decay conserves the lepton number. In $\beta^-$ decay, the atomic number of the element increases which is accompanied by the release of an electron antineutrino, $e^-+\bar{v}_e$ .