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xenn [34]
3 years ago
15

How do the biosphere and geosphere interact? give an example

Chemistry
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

Plants

Explanation:

Plants, part of the biosphere, grow in the soil, which is part of the geosphere

hope this helped :)

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If 2.50 mol of copper and 5.50 mol of silver nitrate are available to react by single replacement, identify the limiting reactan
PIT_PIT [208]

Answer:

The limiting reactant is the copper

Explanation:

Moles are the standard unit at which you can compare values, and copper has fewer moles than Silver nitrate, so in a reaction, it would be used up first, leaving Silver nitrate in excess.

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Helpppp pleaseee for 15 points
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Answer:

<h3><em>What is filteration</em><em>:</em></h3>

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<h3><em>When</em><em> </em><em>do</em><em> </em><em>we</em><em> </em><em>use</em><em> </em><em>it</em><em>:</em></h3>

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4 0
3 years ago
What is a face atom?
gavmur [86]

Answer:

Explanation:

An atom on a face is shared by two unit cells, so only half of the atom belongs to each of these cells. An atom on an edge is shared by four unit cells, and an atom on a corner is shared by eight unit cells.

3 0
3 years ago
CaCl2 can be melted to produce calcium metal and give off chlorine gas. The equation for this is CaCl2(l) Ca(s) + Cl2(g). If 277
Bess [88]

Answer:

82g

Explanation:

CaCl2 : Ca 111g : 40g

227.45 : xg

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4 0
3 years ago
Read 2 more answers
How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge
Ivan

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:-  The balanced equation for the combustion of ethane is:

2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})

= 126 L O_2

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

105.0LO_2(\frac{4LCO_2}{7L O_2})

= 60.0 L CO_2

Similarly, let's calculate the volume of water vapors formed:

105.0L O_2(\frac{6L H_2O}{7L O_2})

= 90.0 L H_2O

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

105.0LO_2(\frac{2LC_2H_6}{7LO_2})

= 30.0 L C_2H_6

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

5 0
3 years ago
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