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3 years ago

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Chemistry

1 answer:

Mila [183]3 years ago

7 0

Answer:

Explanation:

Half life of Nobelium-253 is 97 seconds . That means after every 97 seconds half of the Nobelium amount will be disintegrated .

Time taken in bringing the sample to laboratory = 291 seconds

291 second = 291 / 97 half life

n = 3

N = $N_0 (\frac{1}{2})^n$

N₀ is original mass , N is mass after n number of half life.

N = 5 mg x $(\frac{1}{2})^3$

= .625 mg

Only 0.625 mg of Nobelium-253 will be left .

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A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

$-w=\frac{dm}{dt}$

w is the mass flow

m is the mass of salt

$-v*C=\frac{dm}{dt}$

v is the volume flow

C is the concentration

$C=\frac{m}{V+(6-3)*L/min*t}$

$-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}$

$-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}$

$-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)$

$-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)$

$m=90kg*[2000L/(2000L+3*L/min*t)]$

a) Initially: t=0

$m=90kg*[2000L/(2000L+3*L/min*0)]=90kg$

b) t=210 min (3.5 hr)

$m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg$

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0

3 years ago

(100 POINTS AND BRAINLYEST)

Blababa [14]

Answer:

Sea level changes due to change in temperature leading to thermal expansion

6 0

3 years ago

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In the covalent bond between hydrogen and chlorine in the HCl molecule, where will the majority of shared electrons be found?

Mrac [35]

Nearer to the chlorine as it has a greater electronegativity

5 0

2 years ago

Read 2 more answers

Determine whether each melting point observation corresponds to a pure sample of a single compound or to an impure sample with m

ZanzabumX [31]

Answer:

Wide melting point range - impure sample with multiple compounds

Experimental melting point is close to literature value - pure sample of a single compound

Experimental melting point is below literature value - impure sample with multiple compounds

Narrow melting point range - pure sample of a single compound

Explanation:

The melting point of substances are easily obtainable from literature such as the CRC Handbook of Physics and Chemistry.

A single pure substance is always observed to melt within a narrow temperature range. This melting temperature is always very close to the melting point recorded in literature for the pure compound.

However, an impure sample with multiple compounds will melt over a wide temperature range. We also have to recall that impurities lower the melting point of a pure substance. Hence, the experimental melting point of an impure sample with multiple compounds is always below the literature value.

6 0

2 years ago

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.74 g of water at 52.2 oC in an insulated container. clead = 0.128

QveST [7]

Answer:

Explanation:

Hello, in this case, the lead is catching heat and the water losing it, that's why the heat relation ship is (D is for Δ):

$DH_{lead}=-DH_{water}$

Now, by stating the heat capacity definition:

$m_{Pb}C_{Pb}*(T_{eq}-T_{lead}=-m_{H_2O}C_{H_2O}*(T_{eq}-T_{H_2O})\\$

Solving for the equilibrium temperature:

$T_{eq}=\frac{m_{Pb}C_{Pb}T_{Pb}+m_{H_2O}C_{H_2O}T_{H_2O}}{m_{Pb}C_{Pb}+m_{H_2O}C_{H_2O}} \\\\T_{eq}=\frac{2.04g*0.128J/(g^oC)*10.8^oC+7.74g*4.18J/(g^oC)*52.2^oC}{2.04g*0.128J/(g^oC)+7.74g*4.18J/(g^oC)} \\\\T_{eq}=51.87^oC$

Which is very close to the water's temperature since the lead's both mass and head capacity are lower than those for water.

Best regards.

5 0

2 years ago