A gas changes directly to a solid during ....fusion
saponification
depositon
decomposition

Answers

Answer 1

Answer: The answer is Deposition.

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When does phenolphthalein turn pink?a) in the presence of a base

b) in the presence of an acid

c) when it is in a neutral solution

d) when it is reacting with a metal

Answers

Answer is: a) in the presence of a base.

Phenolphthalein is colorless in acidic solutions and pink in basic solutions.

Acid-base indicators are usually weak acids or bases and they are chemical detectors for hydrogen or hydronium cations.

Example for acid-base indicator is phenolphthalein (molecular formula C₂₀H₁₄O₄).

When solution turns phenolphthalein pink, it means it is basic (pH>7).

Final answer:

Phenolphthalein turns pink in the presence of a base.

Explanation:

Phenolphthalein turns pink in the presence of a base. It is a pH indicator that changes color depending on the acidity or alkalinity of a solution. In an acidic solution, phenolphthalein remains colorless, while in a basic solution, it turns pink.

Learn more about Phenolphthalein turning pink here:

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2. A sample of iron has the dimensions of 2.0 cm x 3.0 cm x2.0 cm. It has a mass of 94.0 grams. What is the density of
iron? (8 points)

Answers

Answer:

first change cm to m and g to kg so you will find the volume first which is 0.02 by 0.03 by 0.02 your answer is 0.000012 m . using the formula density=mass/volume take 0.094/ 0.00012 then divide

Which element is classified as an alkali metal?

Answers

Explanation:

All the elements of group 1 in the periodic table are known as alkali metals. These are lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (fr).

Characteristics of alkali metals are as follows.

Alkali metals are soft and can be cut with a knife.
They are good conductors of heat and electricity.
They have only 1 electron in their outermost sub-shell.
They are very reactive.
Alkali metals are ionic in nature.

The alkali metals are a group (column) in the periodic table consisting of the chemical elements lithium (Li), sodium (Na), potassium (K), [note 1] rubidium (Rb), caesium (Cs), [note 2] and francium (Fr).

Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2. What is the displacement of Andy’s car after 30.0 seconds?A.
1200 meters
B.
2400 meters
C.
3600 meters
D.
4800 meters
E.
7200 meters

Answers

The answer is B
8m/s
Your going to multiply 8 and 30 together which gives you 240.

When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, provided thatA) it is a reactant in one intermediate reaction and a catalyst in the other reaction.
B) it is a product in one intermediate reaction and a catalyst in the other reaction.
C) it is a reactant in one intermediate reaction and a product in the other reaction.
D) it is a reactant in both of the intermediate reactions.

Answers

Answer: Option (C) is the correct option.

Explanation:

When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, provided that it is a reactant in one intermediate reaction and a product in the other reaction.

For example,

$2Al(s) + 2KOH(aq) + 6H_(2)O\rightarrow 2KAl(OH)_(4)(aq)+ 3H_(2)(g)$ ....(1)

$2KAl(OH)_(4)(aq) + 4H_(2)SO_(4)(aq) + 6H_(2)O(l)\rightarrow 2KAl(SO_(4))_(2)(aq)+ 12H_(2)O(s)$ .........(2)

Cancelling the common species in both the equations as follows.

$2Al(s) + 2KOH(aq) + 6H_(2)O\rightarrow \not{2KAl(OH)_(4)(aq)}+ 3H_(2)(g)$

$\not{2KAl(OH)_(4)(aq)}+ 4H_(2)SO_(4)(aq) + 6H_(2)O(l)\rightarrow 2KAl(SO_(4))_(2)(aq)+ 12H_(2)O(s)$

Therefore, on addition we get the equation as follows.

$2Al(s)+ 2KOH(aq) + 4H_(2)SO_(4)(aq) + 22H_(2)O(l)\rightarrow 3H_(2)(g)+ 2KAl(SO_(4))_(2) + 12H_(2)O(s)$

The right answer for the question that is being asked and shown above is that: "C) it is a reactant in one intermediate reaction and a product in the other reaction." When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, provided that it is a reactant in one intermediate reaction and a product in the other reaction.

In the important industrial process for producing ammonia (the Haber Process), the overall reaction is:N2(g) + 3H2(g) 2NH3(g) + 100.4 kJ
A yield of NH3 of approximately 98% can be obtained at 200°C and 1,000 atmospheres of pressure.
How many grams of Ny must react to form 1.7 grams of ammonia, NH3?
0.052 g
1.49
0.00589
2.8 g

Answers

Answer:

Explanation:

Here we have to use stoichiometry.

First of all, we have to calculate the mass of 100% of yield:

1.7 g ------- 98%

X -------- 100%

X = 1.73 g (approximately)

Second, we have to calculate the mass of N2 that is necessary to react to produce the mass of 1.73g of NH3. To do that, we have to use the Molar mass of N2 and NH3 and don't forget the stoichiometric relationship between them.

Molar Mass N2 : 14x2 = 28 g/mol

Molar Mass NH3: 14 + 3 = 17 g/mol

28g (N2) ------- 17x2 (NH3)

X ------------ 1.73 g

X = 1.42 g (approximately)

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In a coffee-cup calorimeter, 100.0 ml of 1.0 m naoh and 100.0 ml of 1.0 m hcl are mixed. both solutions were originally at 24.68c. after the reaction, the final temperature is 31.38c. assuming that all the solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.18 j/8c ? g, calculate the enthalpy change for the neutralization of hcl by naoh

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