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3 years ago

Explain why calcium combines in a different ratio to a fluoride ion versus an oxide ion.

Chemistry

1 answer:

jok3333 [9.3K]3 years ago

5 0

Answer:

This is due the different charges of fluoride and oxide ions.

Explanation:

When calcium reacts it is oxidized to Ca²⁺. In the same way, fluoride ion is reduced to F⁻ and oxide ion to O²⁻.

When these ions are combined, the molecule must be neutral. That means 2 ions of F⁻ are necessaries and just 1 O²⁻ ion will reacts producing:

CaF₂ and CaO.

The different charges of these ions is the reason why calcium will combine in different ratios.

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Mashcka [7]

Answer:

Explanation:

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3 years ago

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How many molecules are in 0.26 mol of CO2

Dominik [7]

1 mole of CO2 = 6.023*10^23 molecules.3 moles of CO2 = 3*6.023*10^23 moleculestherefore, 3 moles of CO2 = 18069*10^20 molecules.

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3 years ago

To determine the molar mass of an unknown organic acid, HA, a 1.056 g sample is titrated with standardized NaOH. Calculate the m

meriva

Explanation:

Equation for the given reaction is as follows.

$HA(aq) + NaOH(aq) \rightarrow NaA(aq) + H_{2}O(l)$

Therefore, moles of NaOH and HA are calculated as follows.

Moles of NaOH = $0.256 M \times 0.03378 L$

= $8.64 \times 10^{-3}$

= 0.00864 mol

Moles of HA = 0.00864

Also, moles = $\frac{\text{weight}}{\text{molecular weight}}$

Molecular weight = $\frac{1.056 g}{0.00864 mol}$

= 122.22 g/mol

Thus, we can conclude that molar mass of given unknown organic acid is 122.22 g/mol.

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3 years ago

What is the molar mass of MnP

igomit [66]

The production of manganese peroxidase (MnP) by Irpex lacteus, purified to electrophoretic homogeneity by acetone precipitation, HiPrep Q and HiPrep Sephacryl S-200 chromatography, was shown to correlate with the decolorization of textile industry wastewater. The MnP was purified 11.0-fold, with an overall yield of 24.3%. The molecular mass of the native enzyme, as determined by gel filtration chromatography, was about 53 kDa. The enzyme was shown to have a molecular mass of 53.2 and 38.3 kDa on SDS-PAGE and MALDI-TOF mass spectrometry, respectively, and an isoelectric point of about 3.7. The enzyme was optimally active at pH 6.0 and between 30 and 40 degrees C. The enzyme efficiently catalyzed the decolorization of various artificial dyes and oxidized Mn (II) to Mn (III) in the presence of H(2)O(2). The absorption spectrum of the enzyme exhibited maxima at 407, 500, and 640 nm. The amino acid sequence of the three tryptic peptides was analyzed by ESI Q-TOF MS/MS spectrometry, and showed low similarity to those of the extracellular peroxidases of other white-rot basidiomycetes.

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4 years ago

What is the molecular mass for a non-electrolyte if 35.0 g of it is dissolved in 45.0 grams of water and the solutions boiling p

adelina 88 [10]

The boiling point of water increases as the amount of impurities dissolved in it increases. For our purposes, we will consider the non-electrolyte to be the dissolved impurity. The change in the boiling point can be calculated using the equation:

$\Delta T_b = i \times K_b \times m$

where $\Delta T_b$ is the change in boiling point, $i$ is the van ‘t Hoff factor (whose value denotes the number of particles each formula unit of the dissolved substance dissociates into in water), $K_b$ is the boiling point elevation constant, and $m$ is the molality (moles of solute/kilogram of solvent) of the solution.

Right off the bat, since we're dealing with a non-electrolyte, the dissolved substance can be assumed not to dissociate in water. So, our van ‘t Hoff factor, $i$ , would be 1 (by contrast, the $i$ for an ionic compound like NaCl would be 2 since, in water, NaCl would dissociate into two particles: one Na⁺ ion and one Cl⁻ ion). We're also given our $K_b$ , which is 0.51 °C/m.

Assuming the normal boiling point of pure water to be 100 °C (a defined value for sig fig purposes), the change in boiling point from having dissolved 35.0 g of the non-electrolyte can be obtained by subtracting 100 °C from the final—elevated—boiling point of 101.25 °C:

$\Delta T_b = 101.25\text{ }^o\text{C} - 100\text{ }^o\text{C} = 1.25\text{ }^o\text{C}$

Now, recall what we're asked to determine: the molecular mass of the dissolved substance. There is one unknown left in the equation: the molality of the solution. Let's first solve for that:

$m = \frac{\Delta T_b}{K_b} = \frac{1.25^\text{ o}\text{C}}{0.51^\text{ o}\text{C}/m} \\ m = 2.45 \text{ mol solute/kg water}.$

Notice that we didn't include the i since its value is 1.

Now, what would happen if we multiplied our molality by the mass of water we've been given? We would be left with the moles of solute. And what are we asked to find? The molecular mass, or the mass per mole. We can accomplish this in two steps. Remember to convert your mass of water to kilograms:

$2.45 \text{ mol solute/kg water} \times 0.045 \text{ kg water} = 0.110 \text{ mol solute.}$

And, finally, we divide the mass of our solute by the number of moles of solute:

$\frac{35.0 \text{ g solute}}{0.110 \text{ mol solute}} = 317.5 \text{ g/mol}$

Our answer to two significant figures (which is the number of sig figs to which our $K_b$ is given) would be 320 g/mol.

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3 years ago