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ryzh [129]
3 years ago
8

If you flip a fair coin 7 times, what is the probability that you will get exactly 2 tails?

Chemistry
2 answers:
Lapatulllka [165]3 years ago
7 0

Answer: 28.571

Explanation: you need to do

100 divided by 7 which is 14.285

then 14.285 times 2 which is 28.571

nexus9112 [7]3 years ago
6 0

Answer:

like 5 since there are 2 sides of a coin

Explanation:

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PLEASE HELPP<br> Tell me everything you know about "balancing the equation" for science
Free_Kalibri [48]
Watch melissa maribel explains it amazingly on her yt channel
3 0
3 years ago
How many are molecules ( or formula) in each sample?
andre [41]

Answer:

  • 4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}
  • 16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

<u>Explanation</u>:

<u>Number of molecules for 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}</u>

\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:

Atomic mass of Na + H + C + 3(O)  = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }

55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)

=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}

<u>Number of molecules for for \left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}</u>

\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.

=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}

8 0
3 years ago
What two molecules were condensed in an aldol condensation to produce (ch3)3cch=chcoch3?
inessss [21]
The given compound is being synthesized by condensing Acetone and Pivaldehyde (Trimethylacetaldehyde).

First Acetone is treated with Base, the base abstracts the mildly acidic proton present at alpha position to carbonyl group. The resulting specie called enolate act as a nucleophile and attacks on highly reactive aldehyde which upon dehydration yields the Aldol Product. The Reaction is as follow,

5 0
3 years ago
What is the molarity of a solution that is made by mixing 35.5 g of Ba(OH)2 in 325 ml of solution?
choli [55]

Answer:

M=0.638M

Explanation:

Hello!

In this case, since the molarity of a solution is calculated by diving the moles of solute by the volume of solution in liters, we first compute the moles of barium hydroxide in 35.5 g as shown below:

n=35.5g Ba(OH)_2*\frac{1molBa(OH)_2}{171.34gBa(OH)_2}\\\\n=0.207mol

Then, the liters of solution:

V=325mL*\frac{1L}{1000mL} =0.325L

Finally, the molarity turns out:

M=\frac{0.207mol}{0.325L}\\\\M=0.638M

Best regards!

5 0
3 years ago
Please Help!
Whitepunk [10]

We can use two equations for this problem.<span>

t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is decay constant.

20 days = 0.693 / λ 
λ   = 0.693 / 20 days        (1) 

Nt = Nο eΛ(-λt)                (2)

Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time taken.
t = 40 days</span>

<span>No = 200 g

From (1) and (2),
Nt =  200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>

</span>Hence, 50.01 grams of isotope will remain after 40 days.

<span>
</span>

3 0
3 years ago
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