Is wind energy consistent or inconsistent

A 200-gram sample of a radioactive nuclide is left to decay for 60 hours. At the end of 60 hours, only 25 grams of the sample is

Goryan [66]

Half life is the time taken by a radioactive isotope to decay by half its original mass.
The original mass is 200 g
Time taken is 60 hours
Final mass is 25 g
Therefore;
Final mass = Original mass × (1/2)^n; where n is the number of half lives.
25 = 200 (1/2)^n
1/8 = (1/2)^n
n = 3
Three half lives = 60 hours
1 half lives = 20 hours
Therefore; the half life of the radioactive nucleus is 20 hours

5 0

2 years ago

how many liters of a 60% antifreeze solution must be added to 8L of a 10% antifreeze solution to produce a 20% antifreeze soluti

Tasya [4]

Answer: 2Liters

Explanation:

The expression used will be :

$M_1V_1+M_2V_2=M_3V_3$

where,

$M_1$ = concentration of first antifreeze= 60%

$M_2$ = concentration of second antifreeze= 10%

$V_1$ = volume of first antifreeze = x L

$V_2$ = volume of second antifreeze = 8 L

$M_3$ = concentration of final antifreeze solution= 20%

$V_3$ = volume of final antifreeze = (x+8) L

Now put all the given values in the above law, we get the volume of antifreeze added

$60\times x+10\times 8=20\times (x+8)$

$x=2L$

Therefore, the volume of 60% antifreeze solution that must be added is 2L

5 0

3 years ago

Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Ac

olga2289 [7]

Answer:

1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of hydrochloric acid = n

Volume of hydrochloric acid solution = 200.0 mL = 0.200 L

Molarity of the hydrochloric acid = 0.089 M

$n=0.089 M\times 0.200 L=0.0178 mol$ of HCL

$HCl(aq)+NaHCO_3(aq)\rightarrow NaCl(aq)+H_2O(l)+CO_2(g)$

According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.

Then 0.0178 moles of HCl wil be neutralized by :

$\frac{1}{1}\times 0.0178 mol=0.0178 mol$ of sodium bicarbonate

Mass of 0.0178 moles of sodium bicarbonate:

0.0178 mol × 72 g/mol = 1.4952 g

1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.

8 0

3 years ago

Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration

elixir [45]

Answer:

8.54

Explanation:

At equivalence point :

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052 = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :

pH = 1/2 [ pKw + pKa + log c ]

Ka for acetic acid = 1.75 X 10⁻⁵

so pKa = -log (1.75 X 10⁻⁵) = 4.74

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169

so log c = log 0.02169 = -1.66

putting the values....

pH = 1/2 [14 + 4.74 - 1.66 ]

pH = 1/2 [ 17.08] = 8.54

6 0

3 years ago

Read 2 more answers

Air trapped in a cylinder fitted with a piston occupies 145 mL at 1.08 atm

pav-90 [236]

Answer: 0.0014 atm

Explanation:

Given that,

Original pressure of air (P1) = 1.08 atm

Original volume of air (T1) = 145mL

[Convert 145mL to liters

If 1000mL = 1l

145mL = 145/1000 = 0.145L]

New volume of air (V2) = 111L

New pressure of air (P2) = ?

Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law

P1V1 = P2V2

1.08 atm x 0.145L = P2 x 111L

0.1566 atm•L = 111L•P2

Divide both sides by 111L

0.1566 atm•L/111L = 111L•P2/111L

0.0014 atm = P2

Thus, the new pressure of air when the volume is decreased to 111 L is 0.0014 atm

7 0

3 years ago