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krek1111 [17]
3 years ago
13

What Group is this element in? (Urgent plz help)

Chemistry
1 answer:
Zepler [3.9K]3 years ago
5 0
Since this is Argon it would be a noble gas
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How many elements and atoms are in marble
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Explanation:

Scientists know that there are 6x1023 molecules in a mole - so we have about 0.5x1023 molecules in our marble…and since every silicon dioxide molecule has one atom of silicon and two of oxygen, we have a grand total of 1.5x1023 atoms. That's 150,000,000,000,000,000,000,000 atoms

7 0
3 years ago
A compound with the formula C6H14 was reacted with Cl2/light to give a mixture of 5 different monochlorinated products (not incl
allochka39001 [22]

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Hexane.

Explanation:

Hello!

In this case, since the general reaction of the compound C4H14 with chlorine is:

C_6H_{14}+Cl_2\rightarrow C_6H_{13}Cl+HCl

Which stands for a substitution chemical reaction in which one chlorine is able to replace one hydrogen and therefore hydrogen chloride gives off; we infer that the initial compound, C4H14, shows off the C_nH_{2n+2} formula characteristic of alkanes; in such a way, as it has six carbon atoms, we infer it is hexane.

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3 years ago
What is the ratio of ions of aluminium fluoride component i.e. Aluminium ions to fluorine ions
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Al:F3​:: 3:19.

Explanation:

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The Arrhenius theory explains how the concentration of H
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Answer:true

Explanation:

7 0
2 years ago
sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction
prisoha [69]

Answer:

2.76 × 10⁻¹¹  

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

                         C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹:       0         0         -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

\text{The relationship between $\Delta G^{\circ}$ and K  is}\\\Delta G^{\circ} = -RT \ln K

T = (25.0 + 273.15) K = 298.15 K

\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}

3 0
3 years ago
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