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nata0808 [166]
2 years ago
7

The beakers on the left and right both contain the same amount of HCL (liquid) and the same mass of calcium carbonate (solid). I

n which beaker will the reaction rate be the greatest?
A) The reaction rate will be highest in beaker 1 because there is a larger surface area of calcium carbonate.

B) The reaction rate will be highest in beaker 2 because there is a larger surface area of calcium carbonate.

C) The reaction rate will be highest in beaker 1 because there is a smaller surface area of calcium carbonate.

D) The reaction rate will be highest in beaker 2 because there is a smaller surface area of calcium carbonate

Chemistry
1 answer:
ruslelena [56]2 years ago
7 0

Answer:

B

Explanation:

Smaller pieces results in a larger surface area, therefore the answer is beaker B.

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Stolb23 [73]

Density = 0.186 g/cm^3

Step 1. Calculate the <em>volume</em> of the block

V = lwh = 9.8 cm × 5.6 cm × 4.2 cm = 230 cm^3

Step 2. Calculate the <em>density </em>

Density = mass/volume = 42.8 g/230 cm^3 = 0.186 g/cm^3

(<em>There is no metal with this low a density</em>.)

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What natural resource did the Mesopotamians use to protect their cities from floods?
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Answer:

mud bricks.

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Consider the steps in coal gasification: C(coal) + H2O(g) → CO(g) + H2(g) ΔH°rxn = 129.7 kJ CO(g) + H2O(g) → CO2(g) + H2(g) ΔH°r
Volgvan

Answer:

The answer to your question is  ΔHrxn = 0 kJ

Explanation:

  Process

1.- Multiply Equation 1 by 2

           2C(coal) + 2H₂O   ⇒   2CO (g) + 2H₂      ΔH rxn = 259.4 kJ

2.- Sum equation 2

           CO(g)     +  H₂O    ⇒    CO₂ (g)  + H₂ (g)   ΔHrxn = -41 kJ

Result

2C + 3H₂O + CO  ⇒  2CO + 3H₂ + CO₂               ΔHrxn =  218.4 kJ

Simplification

        2C + 3H₂O  ⇒   CO + 3H₂  + CO₂    

3.- Sum equation 3

       CO(g) + 3H₂ (g)  ⇒  CH₄ (g)  +  H₂O  (g)       ΔHrxn = -218.4 kJ

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4 0
3 years ago
The neutralization of H3PO4 with KOH is exothermic. H3PO4(aq)+3KOH(aq)⟶3H2O(l)+K3PO4(aq)+173.2 kJ If 60.0 mL of 0.200 M H3PO4 is
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Answer:

Final temperature of solution is 27.48^{0}\textrm{C}

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Let's say final temperature of mixture is T ^{0}\textrm{C}

So, \Delta T_{mixture}=(T-23.43)^{0}\textrm{C}

Here m_{mixture}=135.6 g and C_{mixture}=3.78J/(g.^{0}\textrm{C})

Moles of H_{3}PO_{4} are added = \frac{0.200}{1000}\times 60.0moles = 0.012 moles

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or, T = 27.48^{0}\textrm{C}

So, final temperature of solution is 27.48^{0}\textrm{C}

3 0
3 years ago
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