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uysha [10]
3 years ago
8

Which of the following elements with exhibit the greatest shielding effect

Chemistry
1 answer:
Dafna11 [192]3 years ago
8 0

Rubidium would exhibit the greatest shielding [ effect. ]

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If the half life of iridium-182 is 15 years, how much of a 3 gram sample is left after 2 half-lives?
padilas [110]

Answer:

D. 0.75 grams

Explanation:

The data given on the iridium 182 are;

The half life of the iridium 182, t_{(1/2)} = 15 years

The mass of the sample of iridium, N₀ = 3 grams

The amount left, N(t) after two half lives is given as follows;

N(t) = N_0 \left (\dfrac{1}{2} \right )^{\dfrac{t}{t_{1/2}} }

For two half lives, t = 2 × t_{(1/2)}

∴ t = 2 × 15 = 30

\dfrac{t}{t_{(1/2)}} = \dfrac{30}{15} = 2

\therefore N(t) = 3 \times\left (\dfrac{1}{2} \right )^2 = 0.75

∴ The amount left, N(t) = 0.75 grams

4 0
3 years ago
Which element has the same number of valence electrons as hydrogen (H)? A. helium (He). . B. oxygen (O). . C. nitrogen (N). . D.
deff fn [24]
Li is in group 1 so it has same number of valence electrons as hydrogen.
hope it helps.
4 0
3 years ago
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HEY I NEED ALL ANSWERS RIGHT NOW IT'S MISSING AND MY TEACHER ARE GRADING THEM NOW YOU WILL BE MARKED BRAINIEST
emmainna [20.7K]

Molar mass of butane:-

  • 4(12)+10=58u

Moles of Butane:-

  • 100/58=1.7mol

#16

  • 2mols of Butane produce 8mol CO_2
  • 1mol of butane produces 4mol CO_2

Moles of CO_2

  • 4(1.7)=6.8mol

Mass of CO_2:-

  • 44(6.8)=299.2g

#17

  • 2mols of butane need 13mol O_2
  • 1mol of butane needs 6.5mol O_2

Moles of O_2

  • 6.5(1.7)=11.05mol

Mass of O_2

  • 11.05(32)=353.6g
7 0
2 years ago
4. Which of the following statements
Gnesinka [82]

Answer:

The answer is B

Explanation:

6 0
3 years ago
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An order is given to administer methylprednisolone, an anti-inflammatory drug, by IV at a rate of 39 mg every 30 min. The IV bag
Korvikt [17]

Answer:

The Flow rate = 0.0208 mL/min

Explanation:

Data provided:

Rate of dose = 39 mg every 30 min = (39/30) mg/min  = 1.3 mg/min

also,

125mg of methylprednisolone is present in every 2 mL

thus,

concentration = (125/2) mg/mL = 62.5 mg/mL

Now,

The flow rate is given as:

Flow rate = Rate / Concentration

on substituting the respective values, we get

Flow rate = (1.3 mg/min) / (62.5mg/mL)

or

The Flow rate = 0.0208 mL/min

3 0
3 years ago
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