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kotykmax [81]
2 years ago
10

Which processes occur during the second stage of technological design? Check all that apply.

Chemistry
2 answers:
Savatey [412]2 years ago
8 0

Answer: "designing a solution" and "defining criteria of success"

Lena [83]2 years ago
3 0
There is four stages of technological design:1. identify a problem or need.2. design a solution.3. implement, build, test the design.4. determine if the solution met the need.To design a solution is second stage of technological design.
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Calculate the amount (in grams) of kcl present in 75.0 ml of 2.10 m kcl
Lera25 [3.4K]
V = 75 mL = 0,075 L = 0,075 dm³
C = 2.1M
n = ?
---------------
C = n/V
n = C×V
n = 2.1×0,075
n = 0,1575 mol
--------
mKCl: 39+35.5 = 74,5 g/mol

74,5g --------- 1 mol
Xg ------------- 0,1575 mol
X = 74,5×0,1575
X = 11,73375g KCl

:•)
5 0
2 years ago
Predict what will observe in below mention experiment.
Sloan [31]

Predict what will be observed in each experiment below. Rock candy is formed when excess sugar is dissolved in hot water followed by crystallization. A student wants to make two batches of rock candy. He finds an unopened box of "cane sugar" in the pantry. He starts preparing batch A by dissolving sugar in 500 mL of hot water (70 degree C). He keeps adding sugar until no more sugar dissolves in the hot water. He cools the solution to room temperature. He prepares batch B by dissolving sugar in 500 mL of water at room temperature until no more sugar is dissolved. He lets the solution sit at room temperature

a. It is likely that more rock candy will be formed in batch A.

b. It is likely that less rock candy will be formed in batch A.

c. It is likely that no rock candy will be formed in either batch.

d. I need more information to predict which batch is more likely to form rock candy.

Answer: Option A

Explanation:

More rock candy will be formed in the batch A because it is dissolved in hot water and less rock candy will be formed in batch B because the water is not hot.

Formation of the candies require hot water as the solubility of sugar is more in hot water as compared to normal water.

The sugar will be dissolved in water until the time all the space is filled sugar molecules.

Hence, the correct answer is Option A.

6 0
2 years ago
An alpha particle has the same composition as a helium nucleus.<br><br> True<br> False
Schach [20]
Your answer would be True
3 0
1 year ago
Read 2 more answers
How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?
zaharov [31]

Answer:

m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2

Best regards!

7 0
1 year ago
The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &amp;S carvone is measured a
shusha [124]

Answer:

See explanation

Explanation:

% optical purity = specific rotation of mixture/specific rotation of pure enantiomer  * 100/1

specific rotation of mixture = 23°

specific rotation of pure enantiomer = 61°

Hence;

% optical purity = 23/61 * 100 = 38 %

More abundant enantiomer = 100% - 38 % = 62%

Hence the pure  (S) carvone is (-) 62° is the more abundant enantiomer.

Enantiomeric excess = 62 - 50/50 * 100 = 24%

Hence

(R) - carvone  =  38 %

(S) - carvone = 62%

7 0
1 year ago
Read 2 more answers
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