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3 years ago

Henri becquerel and the curies found out that _____.

Chemistry

2 answers:

harkovskaia [24]3 years ago

8 0

atoms were not indivisible and indestructible

natulia [17]3 years ago

5 0

Henri becquerel and the curies found out that atoms were not indivisible and indestructible

Explanation:

The French Physicist, Henri Becquerel, created the appearance of radioactivity. He shared the Nobel Prize in Physics with Marie Curie and Pierre Curie (Marie's husband) in 1903 for their work in radiation. A radionuclide will emit transmission through the process of radioactive sense.

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3 years ago

Read 2 more answers

When 167.2 joules of heat is added to 4.00 grams of water at 10c the resulting temperature is what

Katena32 [7]

$Q=mc \Delta T \\\\
Q=168,2J\\
m=4g\\
c=4.184\frac{J}{g^{o}C}\\\\
\Delta T=\frac{Q}{mc}=\frac{167,2J}{4g*4,184\frac{J}{g^{o}C}}=9,99^{o}C\\\\
\Delta T = T_{f}-T_{i}\\\\
9,99^{o}C=T_{f}-10^{o}C\\\\
T_{f}=9,99^{o}C+10^{o}C=19,99^{o}C\approx20^{o}C$

5 0

3 years ago

The pencil is material or mixture ?please help

disa [49]

Answer:

mixture

Explanation:

6 0

3 years ago

Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: spe

Murljashka [212]

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Heat required to raise the temperature of ice from -20 °C to 0 °C

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 $\frac{J}{g*C}$ and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 $\frac{J}{g*C}$ *20 C and solving: Q=420 J

Heat required to convert 0 °C ice to 0 °C water

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g* $6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}$ = 3.333 kJ= 3,333 J (being kJ=1,000 J)

Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 $\frac{J}{g*C}$ and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 $\frac{J}{g*C}$ *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

3 0

3 years ago

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$