Help--? im so bad at science ugh

Chemistry

2 answers:

sukhopar [10]3 years ago

8 0

Answer:

a volcano is convergent

Explanation:

Anit [1.1K]3 years ago

8 0

A volcano is convergent

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11cm+11.38cm+500.55cm=

Vika [28.1K]

Answer:522.93 centimeters

Explanation:

I calculated the numbers

8 0

3 years ago

An unknown compound has a percent composition of 52.10% potassium, 15.8% carbon, and 32.1% oxygen. The molar mass of the compoun

lianna [129]

To determine the empirical formula and the molecular formula of the compound, we assume a basis of the compound of 100 g. We do as follows:

Mass Moles
K 52.10 52.10/39.10 = 1.33 1.33/1.32 ≈ 1
C 15.8 15.8/12 = 1.32 1.32/1.32 ≈ 1
O 32.1 32.1 / 16 = 2.01 2.01/1.32 ≈ 1.5

The empirical formula would most likely be KCO.
The molecular formula would be K2C2O3.

4 0

4 years ago

Read 2 more answers

For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

nata0808 [166]

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

$\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)$ .

When $\rm H_2$ is added to $\rm H_2C\text{=}CH_2$ (ethene) under heat and with the presence of a catalyst, $\rm H_3C\text{-}CH3$ (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, $\Delta S < 0$ .

The equation for the change in Gibbs Free Energy for a particular reaction is:

$\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})$ .

For a particular reaction, the more negative $\Delta G$ is, the more spontaneous ("favorable") the reaction would be.

Since typically $\Delta S < 0$ for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

$T$ (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of