Ask question

Ask question

All categories

2 years ago

15

How is agriculture depended on industries

1 answer:

gulaghasi [49]2 years ago

3 0

Answer:

The interdependence of agriculture and industry helps the development of both the sectors. The most important aspect of this inter dependence is that the products of one serve as important inputs for the other. Growth of one sector, thus means ample supply of inputs for the other.

Explanation:

You might be interested in

The specific heat of aluminum is 0.214 cal/g.oC. Determine the energy, in calories, necessary to raise the temperature of a 55.5

Natasha2012 [34]

For this problem, we use the formula for sensible heat which is written below:

Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference

Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>

4 0

3 years ago

I just need to know the answer for six please and thank you!

nydimaria [60]

Answer:

Explanation:

3(NH₄)₂Cr₂O₇

8 0

2 years ago

When producing a soluble salt in a reaction between an acid and an alkali, how can you prepare dry solid crystals from the solut

igor_vitrenko [27]

Answer:

aqueous solution

Explanation:

7 0

3 years ago

1. How does adding energy to a solid affect the motion of the particles? 2. How can a gas at room temperature (like oxygen) beco

Oksanka [162]

I am not sure sorry

6 0

3 years ago

For the following reaction, 4.21 grams of hydrogen gas are allowed to react with 10.6 grams of ethylene (C2H4) . hydrogen(g) + e

sergiy2304 [10]

Answer:a) 11.34 g of ethane $(C_2H_6)$ can be formed

b) $C_2H_4$ is the limiting reagent

c) 3.44 g of the excess reagent remains after the reaction is complete

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$

1. $\text{Moles of} H_2=\frac{4.21}{2}=2.10moles$

2. $\text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ require 1 mole of $H_2$

Thus 0.378 moles of $C_2H_4$ will require= $\frac{1}{1}\times 0.378=0.378moles$ of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product and $H_2$ is the excess reagent.

moles of $H_2$ left = (2.10-0.378) = 1.72 moles

mass of $H_2$ left= $moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g$

According to stoichiometry :

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.378 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.378=0.378moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g$

Thus 11.34 g of ethane is formed.

4 0

3 years ago