Please help with science

In the reaction of aluminum bromide with ionized sodium bromide which compound is the Lewis acid

slavikrds [6]

1. What is a property of a base? You should N-O-T taste laboratory chemicals!!! I don't know why textbooks emphasize the taste of acids and bases. But that is the answer.

2. In the reaction of aluminum bromide with ionized sodium bromide, which compound is the Lewis acid? 
What reaction??? 

3 In a neutral solution the [H^+] is ____. 
At 25C a solution is said to be neutral when the hydrogen ion concentration is 1.00x10^-7M. 

4 With solutions of strong acids and strong bases, the word strong refer to ____. 
The strength of electrolytes, including acids and bases, describes the degree to which the substance ionizes. Strong acids and bases ionize completely in water. 

5 Which of the following pairs consists of a weak acid and a strong base? 
a. sulfuric acid, sodium hydroxide == strong, strong 
b.acetic acid, ammonia == weak, weak 
c. acetic acid, sodium hydroxide* == weak, strong 
d. nitric acid, calcium hydroxide == strong, strong 

6. The ionization constant (K^a) of HF is 6.7 x 10^-4. Which of the following is true in a 0.1M solution of this acid? 
a. [HF] is greater than [H^+][F^-].* == Yep 
b. [HF] is less than [H^+][F^-]. == Nope 
c. [HF] is equal to [H^+][F^-]. == if K=1 
d. [HF] is equal to [H^+][F^2-] == nonsense 

7. The process of adding a known amount of solution of known concentration to determine the concentration of another solution is called ____. 
The process of finding the concentration of an acid or base by neutralizing it with a known concentration of a known volume is a titration.

8 0

3 years ago

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Which of the following is the best name for N,O?

Vinil7 [7]

Sodium is the best na

6 0

3 years ago

Suppose 0.795 g of sodium iodide is dissolved in 100. mL of a 39.0 m M aqueous solution of silver nitrate.

lubasha [3.4K]

Answer:

The final molarity of iodide anion is 0.053 M

Explanation:

Step 1: Data given

Mass of sodium iodide (NaI) = 0.795 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M

The molecular mass of sodium iodide is 149.89 g/mol.

Step 2: The balanced equation

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

Step 3: Calculate number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

Moles NaI = 0.795 grams / 149.89 g/mol

Moles NaI = 0.0053 moles

For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) + I-(aq)

Step 4: Calculate iodide ions

For 1 mole NaI, we have 1 mole of I-

For 0.0053 moles of NaI we'll have 0.0053 moles I-

Step 5: Calculate molarity of iodide ion

Molarity = moles I- / volume

Molarity I- = 0.0053 moles / 0.1 L

Molarity I- = 0.053 M

The final molarity of iodide anion is 0.053 M

5 0

3 years ago

There are three known forms of uranium. These forms are called _____ of each other. a. isotopes b. copies c. triplicates d. mole

KIM [24]

The answer is isotopes. Isotopes are different kind of atoms of the same element. Elements are identified by the number of protons in the nucleus (atomic number), so all the aoms of uranium have the same number of protons. When one element can have atoms with different number of neutrons (which is pretty normal) the different forms are called isotopes. So, isotopes are different versions of the same element which differentiate from each other in the number of neutrons.

7 0

3 years ago

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2. The following reaction takes place in a galvanic cell: 2Al(s) + 3Cu² (aq) → 2Al(aq) + 2Cu(s)

pychu [463]

The standard cell potential is obtained from the equation of the reaction.

<h3>What is the standard cell potential?</h3>

The standard cell potential is obtained from the equation of the reaction. Now we have the overall cell equation as from the half reactions as;

Oxidation half reaction;

Al(s) --->Al³+ (aq) + 3 e

Reduction half equation;

Cu²+ (aq) + 2 e →→ Cu(s)

Standard cell potential Eºcell = 0.34V - (-1.66 V) = 2 V

Now;

ΔG = -nFEcell

ΔG = -(2 * 6 * 96500 * 2)

ΔG = -2316kJ

E = 2 - 0.0592/6 log [0.0025]/[3.5 M]

E = 2 + 0.031

E = 2.031 V

Learn more about standard cell potential:brainly.com/question/14240930

#SPJ1

8 0

2 years ago