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leonid [27]
4 years ago
15

Determine if the following salt is neutral, acidic or basic. If acidic or basic, write the appropriate equilibrium equation for

the acid or base that exists when the salt is dissolved in aqueous solution. If neutral, write only NR. LiNO₃
Chemistry
1 answer:
Nat2105 [25]4 years ago
7 0

Answer:

It is neutral (NR)

Explanation:

Salts are formed when the ionizable hydrogens in an acid is replaced by metallic or ammonium ions from bases. The reaction is known as a neutralization reaction.

The nature of a salt formed from this reaction depends on the nature of the reacting acid and base.

If the reaction is between a strong acid and strong base, the salt produced is a neutral salt.

If the reaction occurs between a strong acid and a weak base, the salt produced is acidic.

If the reaction occurs between a strong base and a weak acid, the salt produced is a basic salt.

Considering the salt above, LiNO3.

On hydrolysis, addition of water, the following products are obtained:

LiNO3 + H2O ----> LiOH + HNO3

The products obtained, LiOH and HNO3 are a strong base and a strong acid respectively. Therefore, the salt, LiNO3, is a neutral salt.

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6.) 50.0 mol H2O<br> ? molecules
8_murik_8 [283]
<h3>Answer:</h3>

3.01 × 10²⁵ molecules H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

50.0 mol H₂O

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u />\displaystyle 50.0 \ mol \ H_2O(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O} ) = 3.011 × 10²⁵ molecules H₂O

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O

5 0
3 years ago
Calcular la normalidad de 1 Kg de sulfuro de aluminio en 5000 ml de solucion.
Troyanec [42]

Calculate the normality of 1 Kg of aluminum sulfide in 5000 ml of solution.

Normality comes out to be 8.11

<h3> Given </h3>
  • Mass of solute: 1000g
  • Volume of solution (V): 5000 ml = 5 liters
  • Equivalent mass of solute (E) = molar mass / n-factor

n-factor for Al_{2}S_{3} is 6 and molar mass is 148g

So, on calculating equivalent mass is equal to 24.66g

FORMULAE of Normality (N) = (Mass of the solute) / (Equivalent mass of the solute (E) × Volume of the solution (V)

                                           N=\frac{1000}{24.66*5}

                                          <u> N=8.11</u>

Therefore, normality of 1 kg aluminum sulfide is 8.11

Learn more about normality here brainly.com/question/25507216

#SPJ10

7 0
2 years ago
How many moles of carbon dioxide are present in 14g of co2
Tomtit [17]

0.32 moles

You just do 14/ 44.01 (the atomic mass of carbon + oxygen's atomic massX2)

8 0
3 years ago
What is an example of a physical change involving iron?
tatiyna
Melting iron would be a physical change

Iron rusting would be a chemical change
6 0
3 years ago
Read 2 more answers
A sample of solid iron is heated with an electrical coil. If 85.7 Joules of energy are added to a 14.3 gram sample initially at
Masja [62]

Answer:

The final temperature of the iron is:- 38.0 °C

Explanation:

The initial temperature = 24.7 °C

Let Final temperature = T °C

Using,

Q = m C ×ΔT

Where,  

Q is the heat absorbed by the iron = 85.7 J

C gas is the specific heat of the iron = 0.45 J/g  °C

m is the mass of iron = 14.3 g

ΔT is the change in temperature. ( T - 24.7 ) °C

Applying the values as:

<u>Q = m C ×ΔT  </u>

85.7 J = 14.3 g × 0.45 J/g  °C  × ( T - 24.7 ) °C

Solving for T, we get that:-

T = 38.0 °C

<u>The final temperature of the iron is:- 38.0 °C</u>

7 0
4 years ago
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