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bezimeni [28]
3 years ago
5

Ethan is investigating a snails response to stimuli. He put the following substances on each corner of a paper towel; syrup,lemo

n juice, vinegar,and water. Then he put a snail in the center of the paper towel to see how it would respond. Which observation is evidence of the snails response to stimuli?
A.the snail likes the water
B.the snail preferred the vinegar
C.the snail moved toward the syrup
D.the snail does not like lemon juice
Chemistry
2 answers:
olga nikolaevna [1]3 years ago
4 0
There doesn't seem to be a clear answer, it is placed in the center when it's substances is on the corners it doesn't give anything else
sergij07 [2.7K]3 years ago
4 0
D.......................
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When sodium peroxide is added to water, sodium hydroxide and oxygen gas is produced.
lukranit [14]

Answer:

The reaction combines the sodium with the hydrogen and oxygen in water to form sodium hydroxide and hydrogen gas, and you get a lot of energy released as heat as well. This heat actually melts any remaining sodium that has not reacted yet, and ignites the hydrogen gas, so you get the bang and the flash.

Explanation:

8 0
2 years ago
Why do the deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer?
rodikova [14]

Answer:

The deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer because Organic compounds that are neither acids or bases do not react with either NaOH or HCl and, therefore remain more soluble in the organic solvent and are not extracted.

Explanation:

4 0
3 years ago
Approximately how many ice cubes must melt to cool 650 milliliters of water from 29°C to 0°C? Assume that each ice cube contains
qwelly [4]

Answer : The number of ice cubes melt must be, 13

Explanation :

First we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}

Density of water = 1.00 g/mL

Volume of water = 650 mL

\text{Mass of water}=1.00g/mL\times 650mL=650g

Now we have to calculate the heat released on cooling.

Heat released on cooling = m\times c\times (T_2-T_1)

where,

m = mass of water = 650 g

c = specific heat capacity of water = 4.18J/g^oC

T_2 = final temperature = 29^oC

T_2 = initial temperature = 0^oC

Now put all the given values in the above expression, we get:

Heat released on cooling = 650g\times 4.18J/g^oC\times (29-0)^oC

Heat released on cooling = 78793 J = 78.793 kJ   (1 J = 0.001 kJ)

As, 1 ice cube contains 1 mole of water.

The heat required for 1 ice cube to melt = 6.02 kJ

Now we have to calculate the number of ice cubes melted.

Number of ice cubes melted = \frac{\text{Total heat}}{\text{Heat for 1 ice cube}}

Number of ice cubes melted = \frac{78.793kJ}{6.02kJ}

Number of ice cubes melted = 13.1 ≈ 13

Therefore, the number of ice cubes melt must be, 13

3 0
3 years ago
Compounds X and Y are both C6H13Cl compounds formed in the radical chlorination of 3-methylpentane. Both X and Y undergo base-pr
zubka84 [21]

Answer:

Y is a 3-chloro-3-methylpentane.

The structure is shown in the figure attached.

Explanation:

The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).

The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.

6 0
3 years ago
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
Vilka [71]

Answer : The correct option is, (C) 1.1

Solution :  Given,

Initial moles of N_2O_4 = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0
3 years ago
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