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Salsk061 [2.6K]

3 years ago

15

Which of the following naming rules would apply to CaCO3?

2 answers:

tatuchka [14]3 years ago

4 0

I believe the correct answer from the choices listed above is the first option. The naming rule that would apply to CaCO3 is should be <span>metal cation + nonmetal anion(ide). The naming would be written as Calcium Carbonate. Hope this answers the question.</span>

Alja [10]3 years ago

3 0

Metal (Ca) Polyatomic anion (CO3). You do not need to use the roman numeral because Ca is in group 1. The answer is metal + polyatomic or Calcium Carbonate

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alexgriva [62]

Answer:

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Explanation:

Step 1. Calculate the <em>molar mass</em> of NaNO₃

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lord [1]

Answer:

2578.99 years

Explanation:

Given that:

100 g of the wood is emitting 1120 β-particles per minute

Also,

1 g of the wood is emitting 11.20 β-particles per minute

Given, Decay rate = 15.3 % per minute per gram

So,

Concentration left can be calculated as:-

C left = $[A_t]=\frac{11.20\ per\ minute}{15.3\ per\ minute\ per\ gram}\times [A_0]= 0.7320[A_0]$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Also, Half life of carbon-14 = 5730 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{5730}\ years^{-1}$

The rate constant, k = 0.000120968 year⁻¹

Time =?

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

So,

$\frac {[A_t]}{[A_0]}=e^{-0.000120968\times t}$

$\frac {0.7320[A_0]}{[A_0]}=e^{-0.000120968\times t}$

$0.7320=e^{-0.000120968\times t}$

$ln\ 0.7320=-0.000120968\times t$

<u>t = 2578.99 years</u>

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Answer:

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Explanation:

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