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3 years ago

An example of a change of state is __________. *

2 answers:

7 0

Changes of state are physical changes in matter. They are reversible changes that do not change matter's chemical makeup or chemical properties. For example, when fog changes to water vapor, it is still water and can change back to liquid water again.please mark brainliest if correct

NeX [460]3 years ago

3 0

Answer:

ice cube melting

Explanation:

bcoz it's a chemical change

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A scientist is measuring the rate of an enzyme-catalyzed reaction that produces a red-colored product. as the solution containin

slavikrds [6]

The answer would be A will increase and T <span>will decrease.

The product of this reaction emits red light because it absorbs green and blue light. As the reaction occurs, the concentration of the product increase. This will makes absorbance of green and blue light increases and the solution will become redder.</span>

8 0

3 years ago

Read 2 more answers

Oxidation number of Al(OH)4

Kobotan [32]

Explanation:

The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion. The oxidation number of oxygen in a compound is usually –2. The oxidation state of hydrogen in a compound is usually +1.

The oxidation state of Al in Al(OH)

−

x+4(+1−2)=−1

∴x=+3

The oxidation state of Mn in MnO

y+2(−2)=0

∴y=+4

thank u

8 0

3 years ago

What is the ph of 0.450 m al(no3)3 [ka for al3+(aq) = 1.00x10-5]? express your answer to two decimal places?

White raven [17]

Answer : The pH of the solution is, 2.67

Explanation :

The equilibrium chemical reaction is:

$Al^{3+}+H_2O\rightarrow Al(OH)^{2+}+H^+$

Initial conc. 0.450 0 0

At eqm. (0.450-x) x x

As we are given:

$K_a=1.00\times 10^{-5}$

The expression for equilibrium constant is:

$K_a=\frac{(x)\times (x)}{(0.450-x)}$

Now put all the given values in this expression, we get:

$1.00\times 10^{-5}=\frac{(x)\times (x)}{(0.450-x)}$

$x=0.00212M$

The concentration of $H^+$ = x = 0.00212 M

Now we have to calculate the pH of solution.

$pH=-\log [H^+]$

$pH=-\log (0.00212)$

$pH=2.67$

Therefore, the pH of the solution is, 2.67

8 0

3 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

3 years ago

Does potassium nitrate (KN03) incorporate ionic bonding, covalent bonding, or both? Explain.

ladessa [460]

Answer: $KNO_3$ incorporates both ionic bonding and covalent bonding.

Explanation:

A covalent bond is formed when an element shares its valence electron with another element. This bond is formed between two non metals.

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element and the element which accepts the electrons is known as electronegative element. This bond is formed between a metal and an non-metal.

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here potassium is having an oxidation state of +1 called as $K^{+}$ cation and nitrate $NO_3^{-}$ is an anion with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $KNO_3$ . $NO_3^-$ is formed by sharing of electrons between two non metals nitrogen and oxygen.

Thus $KNO_3$ incorporates both ionic bonding and covalent bonding.

4 0

4 years ago