Answer:
Two methods which help us to conserve water are:
Sprinkler irrigation system: this irrigation has an arrangement of vertical pipes with rotating nozzles on the top. It is more useful in the uneven and sandy land where sufficient water is not available.
Drip irrigation system: this irrigation system has an arrangement of pipes or tubes with very small holes in them to water plants drop by drop just at the base of the root. It is very efficient as water is not wasted at all.
Q6. 3
Q7. 3
Q8. pH
Q18. 3
Q19. 3
Q20. 4
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Inter-molecular forces. The greater the attraction between particles the more energy required to break the forces or "melt" them.
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Answer:
First check to make sure you have a balanced equation
Second use mole ratios to set up what you need to know
Finally convert the moles to grams using the correct molecular weight
I will get you started
According to the reaction above, one mole of Magnesium Nitrate is produced from 2 moles of Nitric Acid
1 mole of Mg(NO3)2/2 moles of HNO3
5 grams of Nitric acid contains
5 g * 1 mole/63 g of Nitric Acid = 0.079365079 moles
Explanation:
The molar concentration of the KI_3 solution is 0.251 mol/L.
<em>Step 1</em>. Write the <em>balanced chemical equation</em>
I_3^(-) + 2S_2O_3^(2-) → 3I^(-) + S_4O_6^(2-)
<em>Step 2</em>. Calculate the <em>moles of S_2O_3^(2-)</em>
Moles of S_2O_3^(2-)
= 27.9 mL S_2O_3^(2-) ×[0.270 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)]
= 7.533 mmol S_2O_3^(2-)
<em>Step 3</em>. Calculate the <em>moles of I_3^(-)
</em>
Moles of I_3^(-) = 7.533 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 3.766 mmol I_3^(-)
<em>Step 4</em>. Calculate the <em>molar concentration of the I_3^(-)
</em>
<em>c</em> = "moles"/"litres" = 3.766 mmol/15.0 mL = 0.251 mol/L
