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3 years ago

Which element will form covalent bonds with chlorine?

1 answer:

5 0

Hydrogen on sharing its one electron with chlorine results in formation of hydrogen chloride molecule and chlorine and hydrogen complete its octet by sharing one electron of each and thus forming a single covalent bond by sharing of two electrons.

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Hot water deep within earth can _______ minerals and transport them someplace else.

Vedmedyk [2.9K]

Answer:

Dissolve

Hope this helps!

Explanation:

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2 years ago

The molecule carbon disulfide (CS2) is nonpolar and has only London dispersion forces between the molecules. Carbon tetrachlorid

stepladder [879]

Answer:

Final Answer: boiling point and high

Explanation:

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4 years ago

Read 2 more answers

A solution contains 0.10 m sodium cyanide and 0.10 m potassium hydroxide. solid zinc acetate is added slowly to this mixture. wh

Brrunno [24]

1) Zn(CH₃COO)₂(s) + 2KOH(aq) = Zn(OH)₂(s) + 2CH₃COOK(aq)

Ksp{Zn(OH)₂}=1.2*10⁻¹⁷

2) Zn(CH₃COO)₂(s) + 2NaCN(aq) = Zn(CN)₂(s) + 2CH₃COONa(aq)

Ksp{Zn(CN)₂}=2.6*10⁻¹³

Ksp{Zn(OH)₂}<Ksp{Zn(CN)₂}

Zn(OH)₂ precipitates first

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3 years ago

A detailed description please...

Fittoniya [83]

Answer:

The diagram on the right. It has increments of 0.1 of a unit and therefore will provide a more precise measurement. The diagram on the right measures 88.4. In terms of the diagram on the left, a decimal answer cannot be determined because the increments are too large. Therefore the diagram on the left is less precise.

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3 years ago

How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$

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3 years ago