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Anton [14]
3 years ago
5

Consider the following reaction:

Chemistry
1 answer:
RUDIKE [14]3 years ago
5 0

Answer:

H+  + Cl-         +    OH-     +     Ca2+     -> Na+   +  Cl-    +   H2O

Explanation:

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The weight percent of concentrated H2SO4, molar mass=980 g/mol, is 960% and its density is 184 g/ml. What is the molarity of con
Elan Coil [88]

Answer:

<em>C</em> H2SO4 = 9.79 M

Explanation:

  • molarity (M) ≡ # dissolved species / V sln
  • H2SO4 ↔ H3O+  +  SO4-

∴ %w/w H2SO4 = 960% = g H2SO4 / g sln * 100

⇒ 9.6 = g H2SO4 / g sln

calculation base: 1000 g sln

⇒ g H2SO4 = 9600g

⇒<em> </em>mol<em> </em>H2SO4 =<em> </em>9600 g H2SO4 * ( mol H2SO4/ 980g H2SO4 ) = 9.796  mol H2SO4

⇒ V sln = 1000g sln / 1000g/L = 1 L sln

∴ ρ H20 ≅ 1000 Kg/m³ = 1000 g/L

⇒ <em>C</em> H2SO4 = 9.796 mol H2SO4 / 1 L sln

⇒ <em>C</em> H2SO4 = 9.796 M

8 0
4 years ago
A crime lab received a 235-gram sample. The sample had a molecular mass of 128.1 grams and the empirical formula is CH2O. How ma
kvv77 [185]

Explanation:

Given parameters:

Mass of sample = 235g

Molecular mass of sample = 128.1g

Empirical formula = CH₂O

Unknown:

Mass of each element in the sample = ?

Solution:

To solve this problem, we must know that the empirical formula of any compound is the simplest ratio of the atoms it contains. This is not the true formula of the compound.

      Molecular formula = (Empirical formula)ₙ

Let us find the molecular mass of the sample;

      CH₂O = 12 + 2(1) + 16 = 30g

   

      128.1  = (30)n

          n = 4

The molecular formula of the compound is;    (CH₂O)₄  = C₄H₈O₄

Now to find the grams of each element in the sample;

Express the molecular mass of each element and that of the compound as a fraction and multiply with the given mass;

    For C;

              \frac{4 x 12}{128.1}   x   235\\  = 88.06g

          H; \frac{8 x 1}{128.1}  x  235  = 14.68g

          O: \frac{4 x 16}{128.1} x 235 = 117.41g

learn more:

Mass composition brainly.com/question/3018544

#learnwithBrainly

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T or F? A transformer is a device that can only increase the voltage carried in power lines.
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