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sladkih [1.3K]
2 years ago
7

How many grams are in 0.375 mol LiBr? (MM LiBr=86.85 g/mol)

Chemistry
1 answer:
maks197457 [2]2 years ago
6 0

Answer:

32.57g

Explanation:

mass = mole x molar mass

mass of LiBr = 0.375 x 86.85

mass = 32.57g

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If 25.8 grams of BaO dissolve in enough water to make a 212 gram solution, what is the percent by mass of the solution?
Scorpion4ik [409]
Mass of solute = 25.8 g

mass solution = 212 g

% =(  mass of solute / mass solution ) x 100

% = ( 25.8 / 212 ) x 100

% = 0.122 x 100

= 12.2 %



6 0
3 years ago
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Complete the paragraph about the formation of sodium phosphide.
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<h2>Let us complete it :</h2>

Explanation:

let us study about formation of sodium phosphide

Electrons are transferred from atoms of  sodium to atoms of phosphorous

The sodium atom looses electrons and the phosphorus atoms  

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C. This transfer makes the sodium atoms  acquire positive charge and phosphorous acquire negative charge .

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3 years ago
Which phrase describes the energy of collision?(1 point)
Kruka [31]

The phrase that describes the energy of collision is D. kinetic energy transferred when billiard balls hit each other.

<h3>What is collision?</h3>

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Therefore, kinetic energy transferred when billiard balls hit each other is an example.

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brainly.com/question/11352260

8 0
2 years ago
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Will a precipitate form from Pb(NO3)2(aq) and HCl(aq)?
Rus_ich [418]
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5 0
3 years ago
Given pH = 3.50 Find: [H3O+] and [OH-] Is this acidic, basic or neutral?
lys-0071 [83]

Answer:

Explanation:

Given parameters:

           pH = 3.50

Unknown:

    concentration of [H₃0⁺] = ?

    concentration of [OH⁻] = ?

Solution:

In order to find the unknown, we use some simple expressions which best explains the pH scale and the equilibrium systems of aqueous solutions.

         pH = -log₁₀[H₃O⁺]

         [H₃O⁺] = inverse log₁₀ (-pH) = 10^{-pH} = 10^{-3.5}

          [H₃O⁺] = 3.2 x 10⁻⁴moldm⁻³

       

For the  [OH⁻]:

       we use : pOH = -log₁₀ [OH⁻]

     Recall: pOH + pH = 14

                  pOH = 14 - pH = 14 - 3.5 = 10.5

  Now we plug the value of pOH into pOH = -log₁₀ [OH⁻]

                                   [OH⁻] = 10^{-pOH}

                        [OH⁻] = 10^{-10.5} = 3.2 x 10⁻¹¹moldm⁻³

The solution is acidic as the concentration of H₃0⁺ is more than that of the OH⁻ ions.

                   

8 0
3 years ago
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