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Delicious77 [7]
3 years ago
12

Plz help me thank you so much for helping

Chemistry
2 answers:
malfutka [58]3 years ago
6 0
The answer to it is A
Rina8888 [55]3 years ago
3 0

Answer:

A: matter or mass cannot be created or destroyed

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The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.
Marina86 [1]

Answer:

Therefore it will take 7.66 hours for 80% of the lead decay.

Explanation:

The differential equation for decay is

\frac{dA}{dt}= kA

\Rightarrow \frac{dA}{A}=kdt

Integrating both sides

ln A= kt+c₁

\Rightarrow A= e^{kt+c_1}

\Rightarrow A=Ce^{kt}         [e^{c_1}=C]

The initial condition is A(0)= A₀,

\therefore A_0=Ce^{0.k}

\Rightarrow C=A_0

\therefore A=A_0e^{kt}.........(1)

Given that the Pb_{209}  has half life of 3.3 hours.

For half life A=\frac12 A_0 putting this in equation (1)

\frac12A_0=A_0e^{k\times3.3}

\Rightarrow ln(\frac12)= 3.3k     [taking ln both sides, ln \ e^a=a]

\Rightarrow k=\frac{ln \frac12}{3.3}

⇒k= - 0.21

Now A₀= 1 gram, 80%=0.8

and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram

Now putting the value of k,A and A₀in the equation (1)

\therefore 0.2=1e^{(-0.21)\times t}

\Rightarrow e^{-0.21t}=0.2

\Rightarrow -0.21t= ln(0. 2)

\Rightarrow t= \frac{ln (0.2)}{-0.21}

⇒ t≈7.66

Therefore it will take 7.66 hours for 80% of the lead decay.

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This volcanic rock formed _____ from lava _____ in gas.
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C or a one of those hope it helps !!!!
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WHAT IS AN Ionic bond​
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{\orange { \mathbb{ \huge{QUESTION}}}}

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kossel explained that inert gases r inert due to the electronic configuration which contains 8 electrons in their outermost shell

And other elements loss and gain electrons to form ions and to have electronic configuration same as Noble gas and to get stable

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#SARDAR JI.

6 0
3 years ago
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A certain protein was found to contain 0.292% manganese by mass. Determine the minimum molecular mass of the protein.
Lisa [10]
55u \ \ \ \ \ \ \Rightarrow \ \ \ \ \ 0,292\%\\
x \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ 100\%\\\\
x = \frac{55u*100\%}{0,292\%}\approx18835,62u
4 0
3 years ago
It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis
Norma-Jean [14]

Answer:

a)

The overall  balanced combustion  reaction is written as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

(F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = 23.562

b)

the higher heating values (HHV)_f per unit mass of LPG = 49.9876 MJ/kg

the lower heating values (LHV)_f per unit mass of LPG = 46.4933 MJ/kg

Explanation:

a)

The stoichiometric equation can be expressed as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2  \ + \ bH_2O \ + \ cN_2

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

x = \frac{2a+b}{2} \\ \\ x =  \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95

Equating the coefficient of Nitrogen

c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612

Therefore, The overall  balanced combustion  reaction can now be written as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

Now;  To determine the stoichiometric F/A and A/F ratios; we have:

(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\  (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\  (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562

b)

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8}  = 0.7 \\ \\ \\  m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06  \\  \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6}  = 0.24

Finally; calculating the higher heating values (HHV)_f per unit mass of LPG; we have:

(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg

calculating the lower heating values (LHV)_f per unit mass of LPG; we have:

(LHV)_f = (HHV)_f - \delta H_w \\ \\  (LHV)_f = (HHV)_f  - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f   = 49.9876 \ MJ/kg -  [\frac{3.8*18}{44.2}*2.258 \ MJ/kg]  \\ \\ (LHV)_f = 46.4933 \ M/kg

7 0
3 years ago
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