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8090 [49]
3 years ago
7

is the gradual change in the types of species that live in a community. This can include primary and secondary situations.

Chemistry
2 answers:
Rashid [163]3 years ago
5 0

Answer:

succession

Explanation:

I am taking the test right now. Hope it is right. good luck

Fofino [41]3 years ago
5 0

Answer: the answer is succession

Explanation:

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HELP PLEASE :((( Explain why helium filled balloons deflate over time faster than air filled balloons do?
Shkiper50 [21]

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Because helium has more molecules trying to escape at faster rates

Explanation:

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Naphthalene is soluble in diethyl ether, but it is insoluble in water regardless of the solution pH. Explain why this compound c
Mademuasel [1]

Answer:

Naphthalene is a non polar substance

Explanation:

In chemistry, the principle of like dissolves like is the over aching principle that controls the dissolution of one substance in another.

Naphthalene dissolves in diethyl ether because diethyl ether is a non polar solvent just as naphthalene is a nonpolar substance.

Since water is a polar solvent, it can not dissolves naphthalene at any pH because naphthalene is a nonpolar substance.

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4 years ago
Why is a solution that is 0.10M HCN and 0.10M LiCN considered a good buffer system?
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Because of the common ion effect. Adding a common ion decreases the solubility of the salt which is the intention of a buffer.
8 0
3 years ago
What is skeleton?<br>Write any two function of skeleton .​
KatRina [158]
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7 0
3 years ago
Se sabe que 10 g de calcio reaccionan con 4 g de oxígeno para obtener 14 g de óxido de calcio. Indica la cantidad de óxido de ca
egoroff_w [7]

Answer:

Si se usan 50 gramos de calcio y óxigeno, se obtienen 70 gramos de óxido de calcio.

Explanation:

Hola,

En este caso, la reacción llevada a cabo es:

2Ca+O_2\rightarrow 2CaO

De este modo si asumimos el ejemplo dado, 50 gramos de calcio, cuya masa atómica es 40 g/mol y 50 g de oxígeno, cuya masa atómica como gas diatómico es 32 g/mol, antes de calcular los gramos de óxido de calcio producidos, debemos identificar el reactivo límite. Así, calculamos las moles de calcio disponibles en 50 g:

mol_{Ca}^{disponible}=50gCa*\frac{1molCa}{40gCa} =1.25molCa

Y también las moles de calcio consumidas por los 50 g de oxígeno, utilizando su relación molar 2:1:

mol_{Ca}^{consumidas\ por\ O_2}=50gO_2*\frac{1molO_2}{32gO_2} *\frac{2molCa}{1molO_2} =3.125molCa

Por lo tanto, hay menos calcio disponible que el que consume el oxígeno, por lo que el calcio esel reactivo límite. Ahora, con este, calculamos los gramos de óxido de calcio, cuya masa molar es 56 g/mol, que se producen:

m_{CaO}=1.25molCa*\frac{2molCaO}{2molCa}* \frac{56gCaO}{1molCaO}\\ \\m_{CaO}=70gCaO

Esto quiere decir que de 50 gramos de oxígeno, solo 20 gramos reaccionan para formar 70 gramos de óxido de calcio.

Saludos!

4 0
3 years ago
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