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harina [27]
3 years ago
13

Element A has a half-life of 10 days. A scientist measures out 200 g of this substance. After 30 days has passed, the scientist

reexamines the sample.
How much Element A will remain in the sample?

50 g
25 g
12.5 g
100 g
Chemistry
2 answers:
Monica [59]3 years ago
6 0

Answer:

25 g of an element will remain in  the sample.

Explanation:

Initial mass of an element = N_o = 200 g

Final mass of an elemnt after time ,t = N

t = 30 days

Half life of an element =t_{\frac{1}{2}=10 days

\lambda=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{10 days}=0.0693 days^{-1}

\log[N]=\log[N_o]-\frac{\lambda t}{2.303}

\log[N]=\log[200 g]-\frac{0.0693 days ^{-1}\times 30 days}{2.303}

N = 25.01 49 g ≈ 25 g

25 g of an element will remain in  the sample.

myrzilka [38]3 years ago
4 0
N=N₀*2^(-t/T)

N₀=200 g
T=10 d
t=30 d

N=200*2^(-30/10)=25 g

25 g will remain
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1.6x10^23 lead atoms. Find the weight in grams
Ber [7]
Moles of lead(Pb) = 1.6x10^23/6.02x10^23 = 0.265 moles.

Weight of lead = moles x atomic weight of lead
                         =  0.265x207.2
                         =  54.908 grams.

Hope this helps!
5 0
3 years ago
If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.
Klio2033 [76]

Answer : The final temperature of the mixture is 22.7^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ethanol = 2.3J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ethanol

m_2 = mass of water

\rho_1 = density of ethanol = 0.789 g/mL

\rho_2 = density of water = 1.0 g/mL

V_1 = volume of ethanol = 45.0 mL

V_2 = volume of water = 45.0 mL

T_f = final temperature of mixture = ?

T_1 = initial temperature of ethanol = 9.0^oC

T_2 = initial temperature of water = 28.6^oC

Now put all the given values in the above formula, we get

(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC

T_f=22.7^oC

Therefore, the final temperature of the mixture is 22.7^oC

4 0
3 years ago
When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

How many moles of this process?

Relative atomic mass from a modern periodic table:

  • K: 39.098;
  • N: 14.007;
  • O: 15.999.

Molar mass of \text{KNO}_3:

M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

Number of moles of the process = Number of moles of \text{KNO}_3 dissolved:

\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

What's the enthalpy change of this process?

Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ} for 0.212162\;\text{mol}. By convention, the enthalpy change \Delta H measures the energy change for each mole of a process.

\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

4 0
3 years ago
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Mkey [24]

Answer:

pH of buffer =4.75

Explanation:

The pH of buffer solution is calculated using Henderson Hassalbalch's equation:

pH=pKa+log[\frac{[salt]}{[acid]}

Given:

pKa = 3.75

concentration of acid = concentration of formic acid = 1 M

concentration of salt =  concentration of sodium formate = 10 M

pH=3.75+log[\frac{10}{1}]=3.75+1=4.75

pH of buffer =4.75

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alekssr [168]

Answer:

a i think true.

Explanation:

3 0
3 years ago
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