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Salsk061 [2.6K]
3 years ago
15

Which of the following naming rules would apply to CaCO3?

Chemistry
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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What is the mass if 1.72 moles of sodium nitrate
alexgriva [62]

Answer:

146 g

Explanation:

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

Na =                    22.99

  N =                     14.01

3O = 3 × 16.00 = 48.00

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A wooden tool is found at an archaeological site. Estimate the age of the tool using the following information: A 100 gram sampl
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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3 years ago
The scientist at the weather office predicted the temperatures in the city route route for the next week I will be the weather t
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Answer:

lord

Explanation:

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