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galina1969 [7]
1 year ago
6

Need help with this question ​

Chemistry
1 answer:
Amanda [17]1 year ago
3 0

Answer:

Sugar, Carbon Dioxide, And Water

Explanation:

A chemical substance is a form of matter having constant chemical composition and characteristic properties. Some references add that chemical substance cannot be separated into its constituent elements by physical separation methods, i.e., without breaking chemical bonds. Chemical substances can be simple substances, chemical compounds, or alloys. Chemical elements may or may not be included in the definition, depending on expert viewpoint.

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Read 2 more answers
Balance the equation for the reaction in which sodium oxide reacts with water to form sodium hydroxide.
tensa zangetsu [6.8K]

Answer:

➢ \: Balance \:  the  \: equation \:  for \:  the  \: reaction \\  in \:  which \:  sodium  \: oxide \:  reacts  \\ with  \: water \:  to \:  form  \: sodium \:  hydroxide.

⇒ We have Na2O + H2O --> NaOH. We have 2 sodiums and 2 oxygens and 2 hydrogens on the left side, but only one of each on the right side.

Sodium Oxide + Water → Sodium Hydroxide

⇒ Na2O + H2O → 2NaOH .

Sodium oxide is used in ceramics and glasses. Sodium oxide reacts exothermically with cold water to produce sodium hydroxide solution.

6 0
2 years ago
3% Hydrogen Peroxide has an oral LD50 of 900 mg/kg. Acetic Acid has an oral LD50 of 3310 mg/kg. Which one is more hazardous to c
Lina20 [59]
Answer is: hydrogen peroxide <span>is more hazardous to consume.
</span>Median lethal dose<span>, </span><span>LD50</span><span> (</span>lethal dose, 50%) <span>is a measure of the </span>lethal dose<span> of a </span>toxin. LD50<span> used as a general indicator of a substance's </span>acute toxicity, lower LD50<span> is indicative of increased toxicity (in this case </span>900 mg/kg is lower than <span>3310 mg/kg).</span>
4 0
4 years ago
Using the following thermochemical data, what is the change in enthalpy for the following reaction? Ca(OH)2(aq) + HCl(aq) CaCl2(
sesenic [268]
Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?

CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l),  Δ<span>H = -186 kJ
</span>
CaO(s) + H2O(l) -----> Ca(OH)2(s), Δ<span>H = -65.1 kJ
</span>
1) Ca(OH)2 should be  reactant, so
CaO(s) + H2O(l) -----> Ca(OH)2(s) 
we are going to take as 
 Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ

2) Add 2 following equations
Ca(OH)2(s)---->CaO(s) + H2O(l),                    and ΔH = 65.1 kJ
<span><u>CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ</u>

</span>Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)

Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add  enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ
4 0
4 years ago
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