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luda_lava [24]
3 years ago
10

During which season is the sun highest in the sky?

Chemistry
1 answer:
Scilla [17]3 years ago
8 0
It’s summer ywwwwwwwww
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How many moles are there in 1.09 kg of Al(OH)3?
alina1380 [7]
  question   1
moles  =  mass/molar  mass  of  Al(OH)3
convert   Kg  to  g
that  is  1.09  x  1000=1090g

moles  is  therefore=1090g/78(molar  mass  of Al(OH)3)=  13.974 moles

   question  2
moles=
2.55g/327.2(molar   mass  of  Pb(CO3)2=  7.79  x   10^-3   moles
from  avogadro  constant
1moles=6.02  x10^23  formula  units
what  about  7.79  x  10  ^-3
={(7.79  x  10^-3)moles  x ( 6.02  x10^23)} /1  mole=4.69  x10^21  formula  units
4 0
3 years ago
HELPPPPPP PLZZZz i'll mark brainliest! Infrared (IR) and ultraviolet (UV) light. But above 29°C, the coating on the glass underg
Elena L [17]

Answer:

It goes away

Explanation:

cause the IR light only can see through clean glass.

8 0
2 years ago
Read 2 more answers
What is the acceleration of a jogger who moves in a straight line and increases her speed by 1 mile per hour every 0.5 hours?
Reil [10]
2 miles per every hour ?
6 0
2 years ago
A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.0
andriy [413]

Answer:

a) 16.2 dm^3/mol*h

b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)

Explanation:

We can use the integrated rate equation in order to obtain k.

For the reaction A+ B --> P the reaction rate is written as

Rate = -\frac{dC_A}{dt} = -\frac{dC_B}{dt} = \frac{dC_P}{dt} = kC_AC_B

If C_{A0} and C_{B0} are the inital concentrations and x the concentration reacted at time t, so C_A=C_{A0} -x and C_B=C_{B0} -x and the rate at time t is written as:

-\frac{dx}{dt} =-k(C_{A0} -x)(C_{B0}-x)

Separating variables and integrating

\int\limits^x_0 {\frac{1}{(C_{A0}-x)(C_{B0}-x)} } \, dx = \int\limits^t_0 {k} \, dt

The integral in left side is solved by partial fractions, it can be used integral tables

\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}}{C_{A0}-x}-ln\frac{C_{B0}}{C_{B0}-x}) =kt

Using logarithm properties (ln x - ln y = ln(x/y))

\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt

Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a C_{B0} of 0.05 mol dm^-3.

Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.

So, the values given are t= 1, C_{A0}=0.075, C_{B0}=0.05, C_{B}=0.02, it implies that the quantity reacted, x, is 0.03 and C_{A}=0.075-x = 0.045. Then, the value of k would be

kt = \frac{1}{0.05-0.075}(ln\frac{0.075*0.02}{0.045*0.05})

k = 16.21 \frac{dm^3}{mol*1h}

b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.

For reactant A, It is solved with the same equation

\frac{1}{C_{B0}-C_{A0}}(ln\frac{C_{A0}C_{B}}{C_{A}C_{B0}}) =kt

but suppossing that C_A= C_{A0}/2=0.0375 so  C_B=C_{B0}- C_{A0}/2=0.0125, k=16.2 and the same initial concentrations. Replacing in the equation

t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.0125}{0.0375*0.05})

t=1.71 h = 1.71*3600 s = 6.1*10^3 s  

For reactant B, C_B= C_{B0}/2=0.025 so  C_A=C_{A0}- C_{B0}/2=0.05, k=16.2 and the same initial concentrations. Replacing in the equation

t=\frac{1}{16.2(0.05-0.075)}(ln\frac{0.075*0.025}{0.05*0.05})

t=0.71 h = 0.7*3600 s = 2.5*10^3 s  

Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)

3 0
2 years ago
The diagram show the box for an element in the periodic table . What is the atomic mass of the element shown .
DochEvi [55]

Answer:32.1

Explanation:

8 0
3 years ago
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