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Tems11 [23]
3 years ago
12

What is the concentration of bromide, in ppm, if 115.91 g MgBr2 is dissolved in 1.31 L water.

Chemistry
1 answer:
Ksenya-84 [330]3 years ago
6 0

Answer:- Bromide concentration is 80748 ppm.

Solution:- ppm is milligram per liter. Grams of magnesium bromide are given and we are asked to calculate the ppm concentration of bromide ion, So, let's first calculate the grams of bromide and then divide it by the given volume.

Molar mass of Magnesium bromide = 24.305+2(126.90)=278.105

115.91gMgBr_2(\frac{1molMgBr_2}{278.105gMgBr_2})(\frac{2molBr}{1molMgBr_2})(\frac{126.90gBr}{1molBr})

= 105.78 g Br

Let's convert grams of Br to milligrams:

since, 1 g = 1000 mg

105.78g(\frac{1000mg}{1g})

= 105780 mg

concentration of Br = \frac{105780mg}{1.31L}

= 80748 mg per liter

Since, 1 ppm = 1 mg per liter

So, the bromide concentration is 80748 ppm.

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A student heats 10.52 g of sodium hydrogen carbonate in a crucible until the compound completely decomposes to sodium carbonate
saveliy_v [14]

Answer:

m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3

Explanation:

Hello!

In this case, since the decomposition of sodium hydrogen carbonate is:

2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)

Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:

m_{Na_2CO_3}^{theoretical}=10.52gNaHCO_3*\frac{1molNaHCO_3}{84.01gNaHCO_3}*\frac{1molNa_2CO_3}{2molNaHCO_3}  *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}\\\\ m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3

Best regards!

4 0
3 years ago
When 16.0 g of an unknown compound (a nonelectrolyte) are dis solved in exactly 800. g of water, the solution has a freezing poi
dexar [7]

Answer:

A. 266g/mol

Explanation:

A colligative property of matter is freezing point depression. The formula is:

ΔT = i×Kf×m <em>(1)</em>

Where:

ΔT is change in temperature (0°C - -0,14°C = 0,14°C)i is Van't Hoff factor (1 for a nonelectrolyte dissolved in water), kf is freezing point molar constant of solvent (1,86°Cm⁻¹) and m is molality of the solution (moles of solute per kg of solution). The mass of the solution is 816,0g

Replacing in (1):

0,14°C = 1×1,86°Cm⁻¹× mol Solute / 0,816kg

<em>0,0614 = mol of solute</em>.

As molar mass is defined as grams per mole of substance and the compound weights 16,0g:

16,0g / 0,0614 mol = 261 g/mol ≈ <em>A. 266g/mol</em>

I hope it helps!

3 0
3 years ago
Can someone explain how to do this?
skelet666 [1.2K]
The paper is not clear so please ask your problem again with more clear print

8 0
3 years ago
What mass of hcl gas must be added to 1.00 l of a buffer solution that contains [aceticacid]=2.0m and [acetate]=1.0m in order to
Paraphin [41]
PH of acidic buffer = pKa + log [CH₃COONa - HCl] / [CH₃COOH + HCl]
pKa of CH₃COOH = 4.74
Concentration of acetic acid in buffer = 2.0 M
Concentration of sodium acetate = 1.0 M
Concentration of HCl must add = x
pH = 4.74 + log (1-x) / (2+x) = 4.11
x = concentration of HCl must be added = 0.43 M
number of moles of HCl = M * V = 0.43 * 1 = 0.43 mol
mass of HCl must be added = 0.43 * 36.5 = 15.7 g
 
3 0
3 years ago
If an object has a mass of 3500 mg what is the mass in kg? Show detailed work
dangina [55]
The answer is 0.0035

In order to figure out kg, you have to know that 1 milligram is equal to 1/100000 kilograms. You take the amount of mg's that you have, which is 3500, and divide it by 1000000 which will give you your answer.

3500/1000000= 0.0035 kg
7 0
3 years ago
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