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4 years ago

How many nanoseconds are there in one minute?

Chemistry

1 answer:

umka21 [38]4 years ago

3 0

1 minute equals 6e+10 nanoseconds.
I hope this helped! :>

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20 POINTS TO ANYONE WHO CAN ANSWER THIS QUESTION!!!!!!!!!!!!!!!!!!!!!!!

Grace [21]

Answer:

touch, smell, hear, taste, and sight

Explanation:

6 0

4 years ago

Read 2 more answers

Draw the structure of the bromohydrin formed when (Z)-3-hexene reacts with Br2/H2O. Use the wedge/hash bond tools to indicate st

Ipatiy [6.2K]

Answer:

(3R,4R)-4-bromohexan-3-ol

Explanation:

In this case, we have reaction called halohydrin formation. This is a markovnikov reaction with anti configuration. Therefore the halogen in this case "Br" and the "OH" must have different configurations. Additionally, in this molecule both carbons have the same substitution, so the "OH" can go in any carbon.

Finally, in the product we will have chiral carbons, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)

I hope it helps!

5 0

4 years ago

Consider the dissolution of MnS in water (Ksp = 3.0 × 10–14). MnS(s) + H2O(l) Mn2+(aq) + HS–(aq) + OH–(aq) How is the solubility

Mademuasel [1]

Answer:

The solubility of MnS will decrease on addition of KOH solution.

Explanation:

As per the equation given:

$MnS(s)+H_{2}O(l) -->Mn^{+2}(aq)+HS^{-}(aq)+OH^{-}(aq)$

On dissolution of MnS in water it gives a basic solution as it gives hydroxide ions.

Now when the we are adding aqueous KOH solution, it will dissociate as:

$KOH(aq)--->K^{+}(aq)+OH^{-}(aq)$

Thus it will further furnish more hydroxide ion,

This will increase the concentration of hydroxide ions (present of product side), the system will try to decrease its concentration by shifting towards reactant side.

Thus the solubility of MnS will decrease on addition of KOH solution.

7 0

3 years ago

Find the molality of this aqueous solution 15.0% by mass kBr (119g/mol).

Vsevolod [243]

Answer:

we will take a 100g sample of this solution for our convenience

so , there is 15 g kBr in this 100g solution

we know that molality is the number if moles of solute / mass of solvent in kg

we need to find the number of moles in 15g kBr

no of moles = 15/119 s

moles = 0.126 moles/ 100g

multiplying both the numerator and the denominator by 10 to get 1 kg in denominator

= 1.26 moles / 1 kg

Hence, the molality is 1.26

would appreciate a brainliest

3 0

4 years ago

Three Balanced equations for Ammonium salt+Alkali=Salt+water+carbon dioxide

arlik [135]

Answer:

NH₄Cl (aq) + NaOH (aq) -> NaCl (aq) + H₂O (l) + NH₃ (g)

(NH₄)₂CO₃ (aq) + 2KOH (aq) -> K₂CO₃ (aq) + 2H₂O (l) + 2NH₃(g)

NH₄NO₃ (aq) + LiOH (aq) -> LiNO₃ (aq) + H₂O (l) + NH₃ (g)

Explanation:

The gas liberated when an alkali reacts with an ammonium salt is NH₃ (ammonia), not CO₂.

4 0

3 years ago