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3 years ago

What type of reaction is

Chemistry

2 answers:

Alex17521 [72]3 years ago

8 0

Answer:

You have a chemical reaction that is occurring

Lisa [10]3 years ago

7 0

It is a single replacement reaction so C
Explanation:
An element and compound (Ca is the element and CuSO4 is the compound) switch to form a new element and compound (CaSO4 is the new compound and Cu is the new element)

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An El Nino often causes the temperature of the Pacific Ocean to increase several degrees which brings huge rainstorms to the coa

Firlakuza [10]

Scientists are able to detect an El Niño<span> event and its effects on the climate through a variety of technological and natural sciences. One of these natural sciences is dendrochronology, or the study of tree rings. Dendrochronologists study the rings of a tree in order to understand climatic conditions during specific time periods. Thin rings often indicate drier seasons while fatter rings indicate rainy seasons. Depending on where the tree is, scientists can see past El </span>Niño<span> events in trees that exhibit signs of much rainier or drier seasons that normal.</span>

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3 years ago

How are acids and bases defined by arrhenius?

Lilit [14]

As defined by Arrhenius: An Arrhenius acid is a substance that dissociates in water to form hydrogen ions (H+). ... An Arrhenius base is a substance that dissociates in water to form hydroxide (OH–) ions. In other words, a base increases the concentration of OH– ions in an aqueous solution.

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4 years ago

The complete combustion of acetic acid, HC2H3O2(I) to form water, H2O(I), and CO2(g), at constant pressure releases 871.7 kJ of

lbvjy [14]

Answer:

HC2H3O2(I) + O2(g) ---> H2O(I) + CO2(g) ΔH= -72.35 kJ

Explanation:

We know that 5.0 g of acetic acid will contain, 5.0g/60 g/mol = 0.083 moles of acetic acid

Now from the reaction equation;

1 mole of acetic acid evolved -871.7 KJ of heat

0.083 moles of acetic acid will evolve 0.083 * -871.7 = -72.35 KJ

For 5.0 g of acetic acid, we can write;

HC2H3O2(I) + O2(g) ---> H2O(I) + CO2(g) ΔH= -72.35 kJ

3 0

3 years ago

Can ya help me out real quick?

Taya2010 [7]

Answer:

302 kj heat is released by lowering the temperature

8 0

3 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

3 years ago