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Colt1911 [192]
3 years ago
5

Help me Quickly please!

Chemistry
1 answer:
Naya [18.7K]3 years ago
8 0
2 a is 8 i believe, not sure but im 88%

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Consider the nuclear equation below. 239/94 Pu—-> X+ 4/2 He. What is X?
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Answer:

X = U (Uranium)

Explanation:

Pu-->235/92 U + 4/2 He

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Which type of radiation has the highest penetrating ability?
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Gamma rays and x- rays have the highest penetrating power
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What is the partial pressure of 0.50 mol Ne gas combined with 1.20 mol Kr gas at a final pressure of 730 torr?
WARRIOR [948]
  The partial pressure  of 0.50  Ne  gas    is   214.71  torr


      calculation

 the partial   pressure  of Ne  = moles  of Ne/total moles  x  final  pressure

 
find  the total  moles  of the air mixture  

that is moles of   Ne  +  moles of K=  0.50 + 1.20  =  1.70 moles
  

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8 0
3 years ago
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You need to prepare at 2.0 mL sample of a diluted drug for injection. The total amount of the drug to be injected in this 2.0 mL
pav-90 [236]

Answer:

a) The concentration of drug in the bottle is 9.8 mg/ml

b) 0.15 ml drug solution + 1.85 ml saline.

c) 4.9 × 10⁻⁵ mol/l

Explanation:

Hi there!

a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.

The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)

If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.

To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline

c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.

Let´s convert it to molarity:

0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l

3 0
3 years ago
The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th
horrorfan [7]

Answer:

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

Explanation:

ClO ( g ) + O_3 ( g )\rightarrow Cl ( g ) + 2 O_2 ( g ),\Delta H^o_{1,rxn} =-122.8 kJ..[1]

2 O_3 ( g )\rightarrow 3O_2 ( g ),\Delta H^o_{2,rxn} = -285.3 kJ..[2]

O_3(g) + Cl(g)\rightarrow ClO (g)+O_2(g),\Delta H^o_{3,rxn}=?..[3]

The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:

[2] - [1] = [3]

\Delta H^o_{3,rxn}=\Delta H^o_{2,rxn}-\Delta H^o_{1,rxn}

=-285.3 kJ-(-122.8 kJ)=162.5 kJ

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

8 0
3 years ago
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