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4 years ago

The shape of a molecule is determined by???

Chemistry

1 answer:

a_sh-v [17]4 years ago

4 0

Answer:

The shape of a molecule is deteremined by

Explanation:

The location of the nuclei and electrons.

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What are the three broad classes of elements

mario62 [17]

The three broad classes of elements are:

Metals
Metalloids
Nonmetals

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4 years ago

What is polarity as used in magnets?

Oduvanchick [21]

In a permanent magnet, a magnetic field is produced by the composite motions of electrons in geometrically aligned atoms.

A magnetic field is characterized by poles called north and south. Magnetic polarity refers to the orientation of these poles in space.

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3 years ago

Read 2 more answers

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .

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3 years ago

Electromagnetic waves consist of

sineoko [7]

Electromagnetic consists of both electrical and magnetic fields so D. :)

3 0

3 years ago

Consider the reaction of CaC2 and water to produce CaCO3 and NH3 according to the reaction CaCN2 + 3H2O → CaCO3 + 2 NH3 . How mu

xz_007 [3.2K]

Answer:

16.27 g of CaCO3 are produced upon reaction of 45 g CaCN2 and 45 g of H2O.

Explanation:

Ca(CN)2 + 3H2O → CaCO3 + 2 NH3

First of all, let's find out the limiting reactant.

Molar mass Ca(CN)2.

Molar mass H2O: 18 g/m

Moles of Ca(CN)2: mass / molar mass

45 g / 92.08 g/m = 0.488 moles

Moles of H2O: mass / molar mass

45g / 18g/m = 2.50 moles

This is my rule of three

1 mol of Ca(CN)2 needs 3 moles of H2O

2.5 moles of Ca(CN)2 needs (2.5 . 3) / 1 = 7.5 moles

I need 7.5 moles of water, but I only have 0.488. Obviously water is the limiting reactant; now we can work on it.

3 moles of water __ makes __ 1 mol of CaCO3

0.488 moles of water __ makes ___ (0.488 . 1) / 3 = 0.163 moles

Molar mass CaCO3 = 100.08 g/m

Molar mass . moles = mass

100.08 g/m . 0.163 moles = 16.27 g

4 0

3 years ago