(8/9-5/6+7/18)*18+5.95*6-1.45*6

(8/9-5/6+7/18)*18+5.95*6-1.45*6

(8/9-5/6+7/18)*18+5.95*6-1.45*6
=8/9*18-5/6*18+7/18 *18+(5.95-1.45)*6
=16-15+7+24
=32

(5 / 9-1 / 6) △ 7 / 18 + 1 / 4

(5 / 9-1 / 6) △ 7 / 18 + 1 / 4
=7 / 18 △ 7 / 18 + 1 / 4
=1 + + 1 / 4
=5 / 4

When the temperature is 20 ℃, the solubility of sodium chloride is 36 g, so it is necessary to prepare saturated solution of table salt?
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Solubility is defined as the mass of solute that can be dissolved by 100g water or other solvents at a certain temperature
20 ℃, 36g, 100g water
For 18G, 100 △ 2 = 50g water is used to obtain 50 + 18 = 68g solution

To prepare 20% salt solution, how many grams of 18% and 23% salt solution are needed?

Let x g of 18% salt solution and y g of 23% salt solution be used. To prepare 20% salt solution, the weight of x g of 18% salt solution plus y g of 23% salt solution is equal to the weight of (x + y) g of 18% salt solution and Y G of 23% salt solution

In order to prepare 100kg salt water with 14.5% concentration, how much 5% salt water is needed, and how many kg salt can be added? (2) if 24% and 5% salt are used
How many kilos of salt water do you need to mix?

(1) Let 5% x kg, then salt (100-x) kg
5%x+(100-x)=100×14.5%
∴x=90,100-x=10
A: 90 kg of 5% salt water, plus 10 kg of salt
(2) Let 24% of ykG, then 5% of (100-y) kg
24%y+5%(100-y)=100×14.5%
∴y=50,100-y=50
A: 50 kg each

Melt 12 grams of salt into 18 grams of water and 9 grams of salt into 3 grams of water. Now you need 14 grams of water with half water and half salt. How many grams of a and B salt water do you need?

The concentration of methylated salt water = 12 / (12 + 18) = 40%, the depth of ethylated salt water = 9 / (9 + 3) = 75%, half of the prepared salt water is 14g, the concentration of methylated salt water = 50%, then ethylated salt water = 14-xg, according to the meaning, X * 40% + (14-x) * 75% = 14 * 50%, the solution is x = 10g, that is, 10g methylated salt water, 14-10 = 4G ethylated salt water

At present, there are 18 kg of 20% salt water. According to the need, how many kg of water should be added to reduce the salt content of salt water to 12%?

Salt = 18 × 20% = 3.6kg
Add water = 3.6 △ 12% - 18 = 12kg

At present, the salt water with 20% salt content is 18 kg. According to the demand, how many kg of water is needed to reduce the salt water and salt content to 12%?

The equation is very simple and clear. Suppose that the water is added by XKG, the salt water is (18 + x) kg now, and the salt in the salt water is unchanged, so the equation (20% * 18) / (18 + x) = 12% can be formulated

At present, there are two kinds of salt water, one containing 12% salt and the other 20% salt. How many kilos of these two kinds of salt water are taken respectively, and they are mixed into 16% salt water 100 kilos
Answer with quadratic equation of two variables

Let's take 12% XKG and 20% ykG
x+y=100(1)
（0.12x+0.2y)/100=0.16（2）
Lianlide:
0.12（100-y)+0.2y=16
That is 12 + 0.08y = 16
The solution is y = 50 and the substitution is x = 50
A: take 12% 50kg, take 20% 50kg

At present, the salt water with 20% salt content is 18 kg. According to the demand, how many kg of water is needed to reduce the salt content of salt water to 12%?

No matter how much water is added, the mass of salt will not change. That is to say, calculate the salt content first! 18 × 0.2 = 3.6kg. The mass of water is 18-3.6 = 14.4kg. If the salt content becomes 12% after adding water, the total mass is 3.6 / 0.12 = 30kg. Then the water content is 30-3.6 = 26.4kg