Reactions can occur because of a precipitation reaction
<h3>Further explanation</h3>
Double-Replacement reactions, happens if there is an ion exchange between two ion compounds in the reactant to form two new ion compounds in the product
To predict whether this reaction can occur or not is one of them, the precipitation reaction. A precipitation reaction occurs if two ionic compounds which are dissolved reacted to produce one of the products of the ion compound does not dissolve. Formation of these precipitating compounds that cause reactions can occur
Chloride solution reacts with Sodium carbonate solution
CaCl₂(aq)+Na₂CO₃(aq)⇒CaCO₃(s)+NaCl(aq)
Because Calcium carbonate, CaCO₃ an insoluble solid , it will precipitates so that a reaction can occur
This is known as Rutherford's gold foil experiment. To align with J.J Thompson's Plum Pudding Model, he expects a beam of alpha particles to just pass through the gold foil undisturbed. However, some were deflected at certain angles. Alpha particles are positive, so it would just go straight through the nucleus, but will deflect if it hits the electrons. <em>Therefore, the answer is: </em><span><em>Particles that struck the nucleus went straight.</em></span>
The correct option is D.
The reactants that combine together to form glucose are carbon dioxide, water and energy from the sun. Six molecules of carbon dioxide combine with six molecules of water in the presence of sunlight to form glucose. The chemical equation for the reaction is given below"
6CO2 + 6H2O + Sunlight = C6H12O6.
Answer:
0.165 mol·L⁻¹
Explanation:
1. Write the <em>chemical equation</em> for the reaction.
HNO₃ + KOH ⟶ KNO₃ + H₂O
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2. Calculate the <em>moles of HNO₃</em>
c = n/V Multiply each side by V and transpose
n = Vc
V = 0.027 86 L
c = 0.1744 mol·L⁻¹ Calculate the moles of HNO₃
Moles of HNO₃ = 0.027 86 × 0.1744
Moles of HNO₃ = 4.859 × 10⁻³ mol HNO₃
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3. Calculate the <em>moles of KOH
</em>
1 mol KOH ≡ 1 mol HNO₃ Calculate the moles of KOH
Moles of KOH = 4.859 × 10⁻³× 1/1
Moles of KOH = 4.859 × 10⁻³ mol KOH
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4. Calculate the <em>molar concentration</em> of the KOH
V = 29.4 mL = 0.0294 L Calculate the concentration
c = 4.859 × 10⁻³/0.0294
c = 0.165 mol·L⁻¹
Answer:
<em>it is a chemical change as there is a change in color</em>
Explanation:
