All categories

2 years ago

5.21 cm is the same distance as

Chemistry

1 answer:

IRISSAK [1]2 years ago

7 0

A snail walking 5 cm

You might be interested in

Explain how the bonding model for sodium metal would differ from the bonding model for sodium chloride, NaCi

Elina [12.6K]

Sodium metal is quite reactive; sodium ions (as in NaCl) are quite unreactive. Cl^- ions are not reactive; they are stingily attracted to positive ions such as Na^+with which they form ionic bonds.

8 0

2 years ago

What is the Gibbs free energy?

Dimas [21]

It's a thermodynamic quantity equal to the enthalpy minus the product of entropy and the absolute temperature

7 0

3 years ago

Read 2 more answers

What astronomical equiptment did hypatia use?

abruzzese [7]

Hypatia used an astrolabe.

8 0

2 years ago

Is the mass of .00566 mol of germanium (Ge) greater than one gram or less than one gram?

Salsk061 [2.6K]

Answer:

Less than one gram

Explanation:

Since there is no whole number before the decimal it means that the number is less than whole meaning it is less than one gram

8 0

1 year ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

Answer: The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

Explanation:

The given chemical equations follows:

Equation 1: $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

Equation 2: $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:

$K_1=9.57\times 10^{-8}$

$K_2=1.46\times 10^{-19}$

Putting values in above equation, we get:

$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

Hence, the value of $K_c$ for the final reaction is $7.16\times 10^{25}$

5 0

2 years ago