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2 years ago

how much hydrogen will be released during the interaction of potassium weighing 8 grams with hydrochloric acid

Chemistry

1 answer:

krek1111 [17]2 years ago

4 0

Answer:

With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below:

2 HCl (aq) + Zn (s) → ZnCl2 (aq) + H2 (g)

Explanation:

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According to the following reaction, how many grams of potassium phosphate will be formed upon the complete reaction of 29.6 gra

Ugo [173]

Answer:

There is 37.36 grams of K3PO4 produced

Explanation:

Step 1: Data given

Mass of H3PO4 = 29.6 grams

KOH is in excess

Molar mass of KOH = 56.11 g/mol

Molar mass of H3PO4 = 97.99 g/mol

Step 2: The balanced equation

3KOH(aq) + H3PO4(aq) ⇔ K3PO4(aq)+3H2O(l)

Step 3: Calculate mass of KOH

Mass KOH = mass KOH / molar mass KOH

Mass KOH = 29.6 grams / 56.11 g/mol

Mass KOH = 0.528 moles

Step 4: Calculate moles of K3PO4

Since KOH is the limiting reactant, We need 3 moles of KOH for each moles of H3PO4, to produce 1 mole of K3PO4 and 3 moles of H2O

For 0.528 moles of KOH we'll have 0.528/3 = 0.176 moles of K3PO4

Step 5: Calculate mass of K3PO4

Mass K3PO4 = moles K3PO4 * molar mass K3PO4

Mass K3PO4 = 0.176 moles * 212.27 g/mol

Mass K3PO4 = 37.36 grams

There is 37.36 grams of K3PO4 produced

8 0

3 years ago

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2 years ago

How many uL are present in 250 mL of H20? (1 uL = 10^-6 Liters)

Vanyuwa [196]

Answer:

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Explanation:

If 1 L = 1000 mL

Then X L = 250 mL

X = (1 × 250) / 1000 = 0.25 L

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if 1 μL = 10⁻⁶ L

then X μL = 0.25 L

X = (1 × 0.25) / 10⁻⁶ =250000 μL

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gulaghasi [49]

Answer:

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Explanation:

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Answer:

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Explanation:

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