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2 years ago

Corrections here: 1. When calcium reacts with water, the temperature changes from 18c to 39c

1 answer:

3 0

When calcium reacts with water, the temperature changes from 18c to 39c is an exothermic reaction as energy is releasing and increasing the temperature.

<h3>What is an exothermic reaction?</h3>

Exothermic reaction are those reactions in which energy is released when a reaction completes.

An example is burning of wood.

Thus, when calcium reacts with water, the temperature changes from 18c to 39c is an exothermic reaction as energy is releasing and increasing the temperature.

Learn more about exothermic reaction

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HElP PLS!! 20 points Which statement best describes a characteristic unique to renewable resources?

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renewable resources are abundant???

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Antoine Lavoisier burned metals in sealed jars. What is your prediction of his experimental results?

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Answer:

The mass of the jar and contents remained the same after the metal was burned.

Explanation:

My prediction about the experimental results is that the mass of the jar and contents remained the same after the metal was burned in the jar.

This is compliance with the law of conservation of mass which states that in a chemical reaction, matter is neither created nor destroyed by bonds are rearranged for new compounds to form.

In compliance with this law, it is expected that the mass of the jar and its content will remain the same before and after the reaction.
No new material was added and no material was removed from the jar.

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3 years ago

A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

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Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

$Molarity=\frac{moles}{\text{Volume of solution(L)}}$

Moles of glucose = $\frac{18.5 g}{180 g/mol}=0.1028 mol$

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = $\frac{0.1028 mol}{0.1 L}=1.028 mol/L$

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = $M_1=1.208 mol$

Volume of the solution taken = $V_1=30.0 mL$

Molarity of the solution after dilution = $M_2$

Volume of the solution after dilution= $V_2=0.500L = 500 mL$

$M_1V_1=M_2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}$

$M_2=0.07248 mol/L$

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

$0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}$

Moles of glucose = $0.07248 mol/L\times 0.1 L=0.007248 mol$

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

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3 years ago

Calculate the pH in titration of a weak acid: What is the pH in titration of formic acid (HCHO2, 0.200 M, 100.0 mL) after the ad

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Answer:

pH = 12.61

Explanation:

First of all, we determine, the milimoles of base:

0.120 M = mmoles / 300 mL

mmoles = 300 mL . 0120 M = 36 mmoles

Now, we determine the milimoles of acid:

0.200 M = mmoles / 100 mL

mmoles = 100 mL . 0.200M = 20 mmoles

This is the neutralization:

HCOOH + OH⁻ ⇄ HCOO⁻ + H₂O

20 mmol 36 mmol 20 mmol

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We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:

NaHCOO → Na⁺ + HCOO⁻

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻ Kb → Kw / Ka = 5.55×10⁻¹¹

These contribution of OH⁻ to the solution is insignificant because the Kb is very small

So: [OH⁻] = 16 mmol / 400 mL → 0.04 M

- log [OH⁻] = pOH → 1.39

pH = 14 - pOH → 12.61

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3 years ago

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egoroff_w [7]

Answer:

When an acid and a base are placed together, they react to neutralize the acid and base properties, producing a salt. The H(+) cation of the acid combines with the OH(-) anion of the base to form water. The compound formed by the cation of the base and the anion of the acid is called a salt.

Explanation:

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