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Basile [38]
4 years ago
3

The reaction between elemental phosphorus p4(s) and o2(g) to make p4o10(s):

Chemistry
1 answer:
Sever21 [200]4 years ago
6 0
<span>The reaction between elemental phosphorus p4(s) and o2(g) is a synthesis type of reaction where two elements combine together to form a compound. Balancing the equation, we get P4(s) + 5 O2(g) = P4O10(s). Here the number of oxygen molecules are balanced on either side by adding 5 molecules of O2 on the product side.</span>
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What is the molecular formula of each compound?<br> (c) Empirical formula HgCl (m = 5 472.1 g/mol)
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The molecular formula of HgCl (m = 5 472.1 g/mol) is Hg2Cl4.

The molecular formula is an expression that defines the number of atoms of each element in one molecule of a compound. It shows the actual number of each atom in a molecule.

<h3>Molecular formula: What is it?</h3>

A chemical formula is a way to communicate information in chemistry about the proportions of atoms that make up a specific chemical compound or molecule. Chemical element symbols, numbers, and occasionally other symbols like parentheses, dashes, brackets, commas, and plus and minus signs are used to represent the chemical elements.

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2 years ago
Which is a problem associated with the use of trees for biomass? I. Potential deforestation II. Burning fossil carbon III. Poten
BigorU [14]

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Both Option (I) and Option (III)

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7 0
4 years ago
What is the molality of a solution that contains 1.34 moles of NaCl in 2.47 kg of solvent
dsp73

The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

<u>Explanation:</u>

Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.

         \text {Molality}=\frac{\text {Moles of solute}}{\text {Mass of solvent}}

As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.

Hence, \text { molality }=\frac{1.34}{2.47}= 0.54 \mathrm{M}

Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

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