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choli [55]
3 years ago
8

hat is the pH of a solution with a hydroxide ion concentration of 10–6M? A. pH = 12 B. pH = 8 C. pH = 6 D. pH = 7

Chemistry
1 answer:
WARRIOR [948]3 years ago
8 0
C=10⁻⁶ mol/L

pH=14-pOH

pOH=-lg[OH⁻]

pH=14+lg10⁻⁶=14-6=8

B. pH = 8

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At elevations higher than sea level the atmospheric pressure is ___ than the atmospheric pressure at sea level
lions [1.4K]

Answer:

lower

Explanation:

The higher the sea level, the lower the atmospheric pressure. This is due to the density of air decreasing as the altitude increases.

5 0
3 years ago
Number of Mg(OH)2 formula units in 7.40 moles of Mg(OH)2.
-Dominant- [34]
The answer is 4.45 × 10²⁴ units.

To calculate this, we will use Avogadro's number which is the number of units (atoms, molecules) in 1 mole of substance:
6.02 × 10²³ units per 1 mole

So, we need a proportion:
If 6.02 × 10²³ units are in 1 mole, how many units will be in 7.40 moles:
6.02 × 10²³ units : 1 mole = x : 7.40 moles

After crossing the products:
1 mole * x =  7.40 moles * 6.02 × 10²³ units
x = 7.40 * 6.02 × 10²³ units
x = 44.5 × 10²³ units = 4.45× 10²⁴ unit
5 0
3 years ago
What is the valence charge of Neon
8_murik_8 [283]
Are electrons located at the outermost shell of an atom
5 0
4 years ago
Read 2 more answers
Dipole-dipole interactions are (weaker than, stronger than, equal to) hydrogen bonds.
kozerog [31]

Answer and Explanation:

Dipole-Dipole interactions are <u>weaker than</u> hydrogen bonds.

Hydrogen bonds are a form of dipole-dipole interactions, being the strongest form of dipole-dipole interactions.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

7 0
3 years ago
Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp
gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200  

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in  j/mol R Ea is

Ea=35.5*1000 j/mol R

ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

brainly.com/question/20305871

#SPJ1

5 0
2 years ago
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