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4 years ago

What is the difference between acute and chronic radiation poisoning

1 answer:

4 0

Acute: Can be repaired by the body using white blood cells.

Chronic: High doses causing internal bleeding and breakdown of cells causing slow death, 90% of the times cannot be fixed by any medicines.

You might be interested in

How many moles are in 100.56g HCI?

DedPeter [7]

Answer:

2.7475 mol≈2.75 mol

Explanation:

M(HCl) = 1.0 + 35.5 = 36.6 g/mol

100.56 g HCl *(1 mol HCl/36.6 g HCl)≈ 2.7475 mol≈2.75 mol

5 0

3 years ago

The activation energy for a given reaction is 56 kj/mol. at what temperature would the rate constant be quadruple what is was at

Alex787 [66]

The rate equation is given as:

k = A e^(- Ea / RT)

Dividing state 1 and state 2:

k1/k2 = e^(- Ea / RT1) / e^(- Ea / RT2)

k1/k2 = e^[- Ea / RT1 - (- Ea / RT2)]

k1/k2 = e^[- Ea / RT1 + Ea / RT2)]

Taking the ln of both sides:

ln (k1/k2) = - Ea / RT1 + Ea / RT2

ln (k1/k2) = - Ea / R (1/T1 - 1/T2)

Since k2 = 4k1, therefore k1/k2 = ¼

ln (1/4) = [- (56,000 J/mol) / (8.314 J / mol K)] (1/273 K – 1/ T2)

2.058 x 10^-4 = 1/273 – 1/T2

T2 = 289.25 K

8 0

3 years ago

A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

Simora [160]

The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

<u>Answer:</u> The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

<u>Explanation:</u>

To calculate the molarity of solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$

Hence, the concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

4 0

3 years ago

HCl(aq) + KOH(aq) → H 2O(ℓ) + KCl(aq)

fredd [130]

Answer:

try a science calculator

Explanation:

5 0

3 years ago

The enthalpy change for converting 10.0 g of ice at -25.0 °C to water at 90.0 °C is __________ kJ. The specific heats of ice, wa

Nina [5.8K]

AH1 = m * c1 * AT1 calculate this for ice (-25C to 0C) AH2 = AHfus(1 mole)=6.01 kJ = 6010 J AH3 = m *c3 * AT3 calculat this for water (0C to 100C) AH4 = AHvap(1mole)=40.67 kJ = 40670 J AH5= m * c5 * AT5 calculate this for steam (100C to 125C)
Sum ---- AH1+AH2+AH3+AH4+AH5
Data m=18g (1mole water)
c1=specific heat ice= 2.09 J/g K c3=specific heat water= 4.18 J/g K c5=specific heat steam= 1.84 J/g K
AT = (Tend - Tinitial) as this is a difference between temperatures it doesn't matter the units Celsius or Kelvin. Kelvin (K)=Celsius (C)+273.15
AT1 = 0C - (-25C)= 25C= 273.15K - 248.15K= 25K AT3= 100C - 0C = 100C= 100K AT5= 125C - 100C= 25C=25K

6 0

3 years ago