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3 years ago

The following are bronsted- lowery acids. Determine what will form when each donates a proton. HI, H2O, NH4+, HNO3

1 answer:

6 0

1) Chemical reaction: HI(aq) → H⁺(aq) + I⁻(aq).
It gives an iodide anion.
2) Chemical reaction: H₂O → H⁺(aq) + OH⁻(aq).
It gives a hydroxide anion.
3) Chemical reaction: NH₄⁺(aq) → H⁺(aq) + NH₃(aq).
It gives ammonia.
4) Chemical reaction: HNO₃(aq) → H⁺(aq) + NO₃⁻(aq).
It gives nitrate anion.

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How can you tell that a chemical change has occurred?

Alik [6]

A. The products of the change are different from the starting

substances.

<u>Explanation:</u>

Whenever there is a physical change it may just affect the phase change but the properties remains the same. Whenever there is an occurrence of a chemical change, it was indicated by some of these things such as,

The products are exactly different from the products.

Chemical properties of these reactants are entirely different from that of the products.

Chemical composition as well as the physical properties of the reactants and the products will change

5 0

3 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

3 0

4 years ago

Given the reaction: 3() + () → 3() + 2()

Scilla [17]

Answer:

i think d is the answer for this

3 0

2 years ago

Mass and energy are conserved Question 13 options: A) only in chemical changes. B) always in physical changes and sometimes in c

Roman55 [17]

Answer:

Explanation:

sometimes it can change chemically but not all the time.

3 0

3 years ago

How many moles in 58 grams of CO3 ?

Romashka [77]

Answer:

60.0089

Explanation:

7 0

3 years ago