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lbvjy [14]
3 years ago
14

The moon has less gravity than we have here on Earth TRUE OR FALSE​

Chemistry
1 answer:
noname [10]3 years ago
6 0

Explanation:

obviously true the gravity of moon is 1/6th of the Earth

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naturally occurring bromine molecules, br2 have masses of 158, 160, and 162. they occur in the relative abundances 25.69%, 49.99
4vir4ik [10]
The average atomic mass of an element can be determined by multiplying the individual masses of the isotopes with their respective relative abundances, and adding them. 

Average atomic mass of Br = 158 amu(0.2569) + 160 amu(0.4999) + 162 amu(0.2431)
Average atomic mass = 159.96 amu

As described in the problem, the relative abundance for Br-79 is 25.69%. This is because 2 atoms of Br is equal to 79*2 = 158 amu. Similarly, the relative abundance of Br-81 is 81*2 = 162, which is 24.31%.
4 0
3 years ago
What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)
blsea [12.9K]
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, (C_{2}H_{5})_{3}N, the reaction will be
(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}

 and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value  = 5.3 x 10^-4

And kb =  \frac{( [( C_{2}H_{5})_{3}NH^{+} ]*  OH^{-} )}{[( C_{2}H_{5})_{3}N]}

 thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

                    =0.00516 M 

Thus, pOH = 2.30 
 pH = 14 - pOH = 11.7
6 0
3 years ago
A 40.0 mL sample of 0.18 M HCI is titrated with 0.36 M CoHsNH2. Dctermine the pH at these points: At the beginning (before base
Whitepunk [10]

Answer:

at the beginning:

pH = 0.745

Explanation:

HCl is a strong acid, so:

  • HCl + H2O  → H3O+  +  Cl-

       0.18 M             0.18        0.18.....equilibrium

before base is added:

∴ [ H3O+ ] ≅ <em>C </em>HCl = 0.18 M

⇒ pH = - Log [ H3O+ ] = - Log ( 0.18 )

⇒ pH = 0.745

8 0
3 years ago
The lab just ran out of 1 M HCl that you need to complete the Benzillic Acid lab. The TA tells you there is 12 M HCl in the fume
Viktor [21]

Answer:

0.83 mL

Explanation:

Given data

  • Initial concentration (C₁): 12 M
  • Initial volume (V₁): ?
  • Final concentration (C₂): 1.0 M
  • Final volume (V₂): 10.0 mL

We can calculate the initial volume of HCl using the dilution rule.

C₁ × V₁ = C₂ × V₂

V₁ = C₂ × V₂ / C₁

V₁ = 1.0 M × 10.0 mL / 12 M

V₁ = 0.83 mL

The required volume of the initial solution is 0.83 mL.

7 0
3 years ago
How hydrogen chloride gas is prepared on labrotary by conc sulpheric acid<br> ​
Citrus2011 [14]

Answer:

It is prepared small amounts of hydrogen cloride for uses in the lab.

It can be  "generated in an HCl generator by dehydrating hydrochloric acid with either sulfuric acid or anhydrous calcium chloride."

4 0
2 years ago
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