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klio [65]
3 years ago
12

What mass is contributed by Br-79?

Chemistry
1 answer:
satela [25.4K]3 years ago
6 0

Answer:

Calculate the mass of BR -79? Bromine has 2 naturally occurring isotopes (Br-79 and Br-81) and has an atomic mass of 79.904 amu

Explanation:

if i get this wrong srry

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Explain how you would change a solution with an original pH of 8.0 to have a pH of<br> 12.0?
pshichka [43]

Answer: Acidity and basicity, proton concentration, the pH scale. You may have noticed that acidic things tend to taste sour, or that some basic. Since the scale is based on pH values, it is logarithmic, meaning that a change of 1 pH unit .

Explanation: PLEASE MARK ME AS BRAINLIEST

6 0
3 years ago
How many grams of nacl are present in 11.00 moles?
Troyanec [42]
Mass of NaCl = moles x molar mass of NaCl 
                       =   11 x 58.443
                       =      642.873 grams.

Hope this helps!
4 0
3 years ago
What is the molarity of a NaOH solution if 48.0 mL of a 0.220 M H2SO4 solution is required to neutralize a 25.0-mL sample of the
Semenov [28]

Answer:

Molarity NaOH = 0.85M (2 sig figs)

Explanation:

48.0ml(0.220M H₂SO₄) + 25ml(Xmolar NaOH)H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O

2(molarity x volume) H₂SO₄ = (molarity x volume) NaOH

2(0.220M x 48.0ml) = 25.0ml x Molarity NaOH

Molarity NaOH = 2(0.220M x 48.0ml)/25.0ml = 0.8448M ≅ 0.85M (2 sig figs)

6 0
3 years ago
The third test you read about uses the apparatus
Vika [28.1K]

Answer:ionic

Explanation: just did it in edgenuity:)

4 0
3 years ago
Read 2 more answers
How much aluminum oxide are produced when 46.5g of Al react with 165.37g of MnO?
solong [7]

Aluminum oxide produced : = 79.152 g

<h3>Further explanation</h3>

Given

46.5g of Al

165.37g of MnO

Required

Aluminum oxide produced

Solution

Reaction

2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)

  • mol Al(Ar = 27 g/mol) :

mol = mass : Ar

mol = 46.5 : 27

mol = 1.722

  • mol MnO(Ar=71 g/mol) :

mol = 165.37 : 71

mol = 2.329

mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776

MnO as a limiting reactant(smaller ratio)

So mol Al₂O₃ based on MnO as a limiting reactant

From equation , mol Al₂O₃ :

= 1/3 x mol MnO

= 1/3 x 2.329

= 0.776

Mass Al₂O₃ (MW=102 g/mol) :

= 0.776 x 102

= 79.152 g

7 0
3 years ago
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