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mezya [45]
3 years ago
14

Lanthanides and Actinides are metals, nonmetals, or metalloids.

Chemistry
1 answer:
Neko [114]3 years ago
7 0
They are the same easily answer metals all the way
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A chemist wishing to do an experiment requiring 47-Ca2+ (half-life = 4.5 days) needs 5.0 μg of the nuclide. What mass of 47-CaCO
NARA [144]

Answer:

5.8μg

Explanation:

According to the rate or decay law:

N/N₀ = exp(-λt)------------------------------- (1)

Where N = Current quantity,  μg

            N₀ = Original quantity, μg

             λ= Decay constant day⁻¹

              t =  time in days

Since the half life is 4.5 days, we can calculate the  λ from (1) by  substituting N/N₀ = 0.5

0.5 = exp (-4.5λ)

ln 0.5  = -4.5λ

-0.6931 = -4.5λ

λ =   -0.6931 /-4.5

  =0.1540 day⁻¹

Substituting into (1)  we have :

N/N₀ = exp(-0.154t)----------------------------- (2)

To receive 5.0 μg of the nuclide with a delivery time of 24 hours or 1 day:

N = 5.0 μg

N₀ = Unknown

t = 1 day

Substituting into (2) we have

[5/N₀]   = exp (-0.154 x 1)

    5/N₀        = 0.8572

N₀  =  5/0.8572

     =    5.8329μg

    ≈     5.8μg

The Chemist must order 5.8μg  of 47-CaCO3

6 0
3 years ago
Please Help Me Answer Numbers 7, 9, 10, and 11!
CaHeK987 [17]
Add a 8 kid and u are just adding
5 0
2 years ago
Jumping on a cemented floor is more pain full than a sandy flour,why​
Anvisha [2.4K]

Answer:

Is soft

Explanation:

because concrete is hard ash so t hink flour would be safer

5 0
3 years ago
At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank
hodyreva [135]

Answer:

\large \boxed{\text{-92 $^{\circ}$C}}

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data  

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}

2. Temperature in degrees Celsius

\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}

8 0
3 years ago
A reduction reaction involves either the addition of hydrogen or removal of:
Marina CMI [18]

Reduction involves the either the addition of hydrogen and removal of oxygen.

<h3>What is reduction?</h3>

Reduction involves the removal of oxygen.

This implies there is a loss of oxygen in reduction.

This can be represented in the extraction of iron from it ores.

Fe₂O₃ + 3CO →  2Fe + 3CO₂

Reduction is also the addition of hydrogen. This implies it is the gain of hydrogen.

For example

CH₃CHO → CH₃CH₂OH

learn more on reduction here: brainly.com/question/9485345

#SPJ12

4 0
2 years ago
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