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tigry1 [53]
3 years ago
14

Science question down below

Chemistry
1 answer:
aliya0001 [1]3 years ago
6 0
Option 1/A (It is the first one)
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How many grams of KNO3 are required to prepare 0.250 L of 0.70 M solution?​
drek231 [11]

Answer:

about 3

Explanation:

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2 years ago
_____________ is an acronym used to refer a family of iron alloys.
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FeNi or NiFe is an acronym used to refer a family of iron alloys.

<h3>What is alloy?</h3>

An alloy is a mixture of chemical elements that contains at least one is a metal.

<h3>Alloy of iron</h3>

The alloy of iron-nickel can be abbreviated as FeNi, which implies iron-nickel.

where;

  • Fe stands for iron
  • Ni stands for Nickel

Thus, FeNi or NiFe is an acronym used to refer a family of iron alloys.

Learn more about iron alloys here: brainly.com/question/24842164

5 0
2 years ago
Which of the following statements is correct?
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D. Blue litmus paper turns red when placed in a base.
7 0
3 years ago
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The temperature of a boiling liquid can rise above its boiling point.<br> a. true<br> b. false
goldenfox [79]
The temperature of a liquid can exceed its boiling point. An example is water. Although at ordinary pressure of 1 atm, the boiling point is 100 degrees, water can still exist in higher temperatures but this time in another state. Superheated steam is the term used for water whose temperature has higher than the boiling point
5 0
3 years ago
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When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°
dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.30^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant =  

m= molality = \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579

8.30^0C=1\times K_f\times 0.579

K_f=14.3^0C/m

Let Mass of solute (KBr) = x g

8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}

x=58.2g

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

7 0
3 years ago
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