1. It is known that the density of 2.00% copper sulfate solution is 1.02g/ml. To prepare 500ml of this solution, how many grams of jarosite are needed? What is the concentration of the substance in the solution? 2. Absorb 560 liters of ammonia with 1 liter of water (standard condition), and the density of ammonia is 0.90 g / cm ^ 3 (1) 560 liters of ammonia (2) The concentration of the substance in the solution obtained by absorbing ammonia with water 3. Heat 26.28 g CaCl2 · nH2O to 200 ℃ to make it lose all crystal water, and the mass becomes 13.32 g. if 43.80 g CaCl2 · nH2O crystal is taken to form 500 ml solution, what is the mass concentration of the solution?

1. It is known that the density of 2.00% copper sulfate solution is 1.02g/ml. To prepare 500ml of this solution, how many grams of jarosite are needed? What is the concentration of the substance in the solution? 2. Absorb 560 liters of ammonia with 1 liter of water (standard condition), and the density of ammonia is 0.90 g / cm ^ 3 (1) 560 liters of ammonia (2) The concentration of the substance in the solution obtained by absorbing ammonia with water 3. Heat 26.28 g CaCl2 · nH2O to 200 ℃ to make it lose all crystal water, and the mass becomes 13.32 g. if 43.80 g CaCl2 · nH2O crystal is taken to form 500 ml solution, what is the mass concentration of the solution?

one
M (CuSO4 solution) = 1.02 * 500 = 510g;
m(cuso4)=510*2.00%=10.2g;
Cuso4 Cuso4.5H2O
160 250
10.2g x
x=250/160*10.2g=15.9375g
n(Cuso4)=10.2/160 mol=0.06375mol;
c(Cuso4)=n/v=0.06375/0.5 mol/l=0.1275mol
two
n=v/22.4=25 mol;
M (ammonia) = 25 * 17g = 425g;
M (water) + m (ammonia) = 1000g + 425g = 1425g;
v=m/p=1425/0.9 ml=1583.3ml=1.60l;
c=n/v=25/1.60 mol/l=15,62mol/l;
three
CaCl2.nH2O CacCL2
111+18n 111
26.28g 13.32g
111+18n/26.28=111/13.32
The solution is n = 6;
The amount of cacl2.6h2o of 43.80g is 0.2mol
C（cacl2)=n/v=0.2/0.5 mol/l=0.4mol/l

The base and area of a triangle and a parallelogram are equal respectively. The area of a triangle is 14 square centimeters and its height is 2.9 centimeters
What is the height of this parallelogram?

14/(14*2/2.9)
=14/(28/2.9)
=14*2.9/28
=1.45 cm

Simple calculation of 3.38 * 0.5 + 4.62 & 2

3.38*0.5+4.62&2?
Is it divided by two?
If so, then
3.38×0.5＋4.62×0.5
＝0.5×（3.38＋4.62）
＝0.5×8＝4

Application of Pythagorean theorem in Mathematics
A robot starts from o point, walks 2m due east to A1 point, 4m due north to A2 point, 6m due west to A3 point, 8m due south to A4 point, 10m due east to A5 point. According to this rule, when the robot goes to A6, draw a figure and calculate the distance between the robot and o point
Drawing I can draw mainly how to find the distance between the robot and point o

The robot's trajectory is right angle change direction, each time increase 2m
When it comes to A6, connect oa6, extend the intersection of OA1 and a5a6 at point B, then: △ oba6 is RT △
Namely: oa6 & # 178; = a6b & # 178; + ob & # 178;
OB=10-(6-2)=6(M)
A6B=12-(8-4)=8)(M)
∴OA6²=8²+6²=80
OA6=√100=10（M)
That is: the distance between the robot and o point is 10m

In ladder ABCD, AD / / BC, ab = DC = ad = 3, BD is perpendicular to CD, the degree of angle DBC and the length of BC are calculated

∫ AB = ad ∈ abd = ∠ ADB ∫ AD / / BC ∈ DBC = ∠ ADB ∫ AD / / BC, ab = DC = ad ∫ trapezoid ABCD is isosceles trapezoid ∠ CBA = ∠ DCB = 2 ∠ ADB ∫ BD vertical CD ∈ bad = ∠ ADC = 90 ° + ∠ ADB ∫ CBA + ∠ DCB + ∠ bad + ∠ ADC = 360 ° 2 ∠ ADB + 2 ∠ ADB + 90 ° + ∠ ADB + 90 ° +} ADB = 360

AB's height is 1.7 * 10 quadratic cm, but a says he is 9cm higher than B. is it possible? Why?

It's possible
In the case of rounding
174cm=1.7*10^2cm
165cm=1.7*10^2cm
The difference is 9cm

If a sector with a central angle of 120 ° and an area of 3 π is taken as the side of the cone, the volume of the cone is______ ．

Let the generatrix of the cone be l, the radius of the bottom be r, ∵ 3 π = 13 π L2 ∵ L = 3, ∵ 120 ° = R3 × 360 °, r = 1, ∵ the height of the cone be 9 − 1 = 22 ∵ the volume of the cone be 13 × π × 12 × 22 = 22 π 3

A piece of rectangular glass is 8 meters in length and half in width. How much is the perimeter and area of this rectangular glass

The width is four
Perimeter = sum of length and width multiplied by 2 = 24m
Area = length times width = 32m ^ 2

Using the rounding method, take the approximate value of 47995483 according to the requirements, and how accurate is it to 10000 bits?

48000000

It takes 90 seconds for a train to pass a 1500 meter long bridge and 55 seconds to pass an 800 meter long tunnel at the same speed!

Let: the length of the train be x meters, (1500 + x) / 90 = (800 + x) / 55. The solution of the equation is: x = 300 meters;
Speed = (1500 + x) / 90 substituting x = 300m, speed = (1500 + 300) / 90 = 20m / S = 72km / h