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Arte-miy333 [17]
3 years ago
5

What is the concentration of bromide, in ppm, if 12.41 g MgBr2 is dissolved in 2.55 L water.

Chemistry
1 answer:
pav-90 [236]3 years ago
7 0

Answer:

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

Explanation:

ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.

In our case we have:

mass of MgBr₂ = 12.41 g

volume of water (which is equal to the final solution volume) = 2.55 L

Now we devise the following reasoning:

if         12.41 g of MgBr₂ are dissolved in 2.55 L of water

then         X g of MgBr₂ are dissolved in 1 L of water

X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂

if in         184 g (1 mole) of MgBr₂ we have 160 g of Br⁻

then in   4.867 g of MgBr₂ we have Y g of Br⁻

Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)

4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)

concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm

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1 year ago
I need help FAST ASAP
kiruha [24]
In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.

1.  Na⁺     ,    Cl⁻    , NaCl   ---> sodium chloride (this is most commonly known as table salt)

2. C⁴⁺       , O²⁻     , CO₂  ---> carbon dioxide

3. Al³+     , Cl⁻       , AlCl₃   ----> aluminum chloride

4. Ca²⁺     , Cl⁻     , CaCl₂    ---> calcium chloride

5. Li⁺        , Br⁻      , LiBr       ---> lithium bromide

6. Mg³⁺     , O²⁻      , Mg₂O₃   ----> magnesium oxide

7. K⁺        , I⁻          , KI   ---> potassium iodide

8. H⁺        , Cl⁻        , HCl  --> hydrogen chloride

9. H⁺        , Br⁻         , HBr ----> hydrogen bromide

10. Na⁺    , Br⁻         , NaBr   ---> sodium bromide
6 0
4 years ago
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victus00 [196]

Answer:

This isotope has 59 electrons giving it a charge of -2.

Explanation:

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6 0
3 years ago
0.350 mol of a solid was dissolved in 260 mL of water at 21.2 oC. After the solid had fully dissolved, the final temperature of
Fittoniya [83]

Answer: Heat of the solution  = mass water × specific heat water × change in temperature

mass water = 260ml (1.00g/ml ) = 260g

specific heat of water = c(water) = 4.184J/ g°C

Heat change of water = final temperature - initial temperature

                                       = 26.5 - 21.2

                                        = 5.3 °C

H = 260 g ( 4.184J/g°C ) (5.3°C) = 5765J

Molar heat = \frac{5765J}{0.350mol}

                    = 16473J/mol

Explanation: finding molar heat requires first to look at  specific heat of water and the change of water temperature

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Answer:

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