Answer:
Gas
Increase the pressure
Explanation:
Let's refer to the attached phase diagram for CO₂ (not to scale).
<em>At -57 °C and 1 atm, carbon dioxide is in which phase?</em>
If we look at the intersection between -57°C and 1 atm, we can see that CO₂ is in the gas phase.
<em>At 10°C and 2 atm carbon dioxide is in the gas phase. From these conditions, how could the gaseous CO₂ be converted into liquid CO₂?</em>
Since at 10°C and 2 atm carbon dioxide is below the triple point, the only way to convert it into liquid is by increasing the pressure (moving up in the vertical direction).
Answer:
Abiotic objects
Ex. bed, table, lamp, chair, blanket, etc.
Abiotic means Non-living.
Balanced equation :
Cu(NO₃)₂(aq) + 2KOH(aq) → Cu(OH)₂(s) + 2KNO₃(aq)
Balancing a chemical equation :
A chemical equation shows us the substances involved in a chemical reaction - the substances that react (reactants) and the substances that are produced (products). In general, a chemical equation looks like this:
Reactant →Product
According to the law of conservation of mass, when a chemical reaction occurs, the mass of the products should be equal to the mass of the reactants. Therefore, the amount of the atoms in each element does not change in the chemical reaction. As a result, the chemical equation that shows the chemical reaction needs to be balanced. A balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.
Learn more about balanced equation :
brainly.com/question/15355912
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Answer:
=zero degrease, 1 atm)? so the volume of an ideal gas is 22.l/mol at STP this, 22l.4Lis probably the most remembered and least useful number in chemistry
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V
