HNO3 is a Bronsted-Lowry acid. What will this substance form when it donates a proton?

1. How many metals, nonmetals and metalloids are there in the periodic table?

sweet-ann [11.9K]

Answer:

There are 110 elements known to us, out of which 92 are naturally occurring, while the rest have been prepared artificially. Elements are further classified into metals, non-metals, and metalloids..

Explanation:

Hope it helps you..

Y-your welcome in advance..

(;ŏ﹏ŏ)(ㆁωㆁ)

8 0

3 years ago

Al2(SO4)3(s) + H2O(l) ---- Al2O3(s) + H2SO4 (aq) calculate enthalpy formation for this reaction. Balance the reaction. Calculate

katen-ka-za [31]

The given chemical equation is:

$Al_{2}(SO_{4})_{3}(aq)+H_{2}O(l)-->Al_{2}O_{3}(aq)+H_{2}SO_{4}(aq)$

On balancing the equation we get,

$Al_{2}(SO_{4})_{3}(aq)+3H_{2}O(l)-->Al_{2}O_{3}(aq)+3H_{2}SO_{4}(aq)$

Calculating enthalpy of formation of this reaction from the standard heats of formation of the products and reactants:

Δ $H_{reaction}^{0}=[H_{f}^{0}(Al_{2}O_{3}(s)) + (3*H_{f}^{0}(H_{2}SO_{4}(aq))] - [H_{f}^{0}(Al_{2}SO_{4}(aq)) + (3*H_{f}^{0}(H_{2}O(l))]$

=[(-1669.8kJ/mol)+ {3* (-909.27 kJ/mol)}]-[(-3442kJ/mol)+{3*(-285.8 kJ/mol)}]

=[(-4397.61kJ/mol)]-[(-4299.4kJ/mol)]

=-98.21kJ/mol

Total enthalpy change when 15 mol of $Al_{2}(SO_{4})_{3}$ reacts will be=

$15 mol Al_{2}(SO_{4})_{3}*\frac{-98.21kJ}{1 molAl_{2}(SO_{4})_{3}} =-1473.15kJ/mol$

4 0

4 years ago

The reaction 2HI → H2 + I2 is second order in [HI] and second order overall. The rate constant of the reaction at 700°C is 1.57

In-s [12.5K]

Answer:

1.135 M.

Explanation:

For the reaction: 2HI → H₂ + I₂,

The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².

To solve this problem, we can use the integral law of second-order reactions:

1/[A] = kt + 1/[A₀],

where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),

t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),

[A₀] is the initial concentration of HI ([A₀] = ?? M).

[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).

∵ 1/[A] = kt + 1/[A₀],

∴ 1/[A₀] = 1/[A] - kt

∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.

∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.

So, the concentration of HI 8 hours earlier = 1.135 M.

8 0

3 years ago

List the six steps of one full cycle of the scientific method post-lab

STatiana [176]

Ask a question
research
create hypothesis
experiment
analyze data
draw conclusions
was your hypothesis correct
share results

7 0

3 years ago

Look.....................................................................

WARRIOR [948]

Answer:

B

Explanation:

I had this question and got it right

8 0

3 years ago