Answers
Answer:
chongus because he's the only good one
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How much heat is required to change the temperature of two cups of water (500 g) from room temperature (25◦C) to boiling? Specific heat of water is c=4.184 J/(g oC) a 78.5 kJ b 15.7 kJ c 157 kJ d 1.57 kJ
I DON'T KNOW THE ORDER!!
Answers
Answer:
for 1 solid its freezing.
for 2 solid and liquid its melting
for 6 liquid to gas its evaporation and for 5 gas to liquid its condensation.
Explanation:
hope this helped :)
Answer:
solid->liquid= melting
liquid->solid= freezing
gas->liquid= consendation
liquid->gas= evaporation
Answers
The amount of heat will be 5230 j.
What is heat?
Heat is a type of energy that is transferred between both the system and its surroundings as a result of temperature variations.
Calculation of heat.
Given data:
Mass = 25.0 g = 0.025 kg
C = 4.184 J/g°C
= 80.0°C
= 30.0°C
Q= ?
By using the formula of heat.
Q = MC ()
Put the value of given data in heat equation.
Q(heat) = 0.025 × 4.184 ( 30 - 80)
Q(heat) = 5230 J.
Therefore, the amount of heat will be 5230 J.
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Answer:
5230 J
Explanation:
m = 25 g = 0,025 kg
c = 4,184 J /(g * °C) = 4184 J /(kg * °C)
= 80 °C
= 30 °C
The formula is Q = c *m * ()
Calculating:
Q = 4184 * 0,025 * (30 - 80) = 5230 J
Note that we get a negative heat (-5230 J). It just means that it is released.
Answers
Answer:
1 litre of 1.0 M NaCl
Explanation:
When an ionic compound dissolves in water, it dissociates into ions. Consider the dissolution of sodium chloride in water;
NaCl(s) ------> Na^+(aq) + Cl^-(aq)
Hence, two solute particles are obtained from each formula unit of NaCl, a greater concentration of NaCl will contain a greater number of sodium an chloride ion particles.
Glucose is a molecular substance and does not dissociate in solution hence it yields a lesser number of particles in solution even at the same concentration as NaCl
Final answer:
The solution with the greatest number of solute particles is 1 litre of 1.0 M NaCl, as ionic compounds dissociate into individual ions, thus providing more particles per litre.
Explanation:
Given the details of the question, the solution that would be expected to contain the greatest number of solute particles would be 1 litre of 1.0 M NaCl. This is because when ionic compounds like sodium chloride are placed in water, they dissociate into individual ions. In the case of NaCl, it splits into two ions, sodium (Na+) and chloride (Cl-). Thus, a 1.0 M solution of NaCl would actually contain 2.0 moles of particles per litre because each formula unit of NaCl gives two particles. Covalently bonded molecules like glucose do not dissociate in solution, therefore, a 1.0 M glucose solution would have 1.0 mole of particles per litre.
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Answers
Answer: Formal Charges: Hydrogen = 0 and Oxygen = +1
Unshared Pair of electrons: Hydrogen = 0 and Oxygen = 2
Explanation:
The attachment below shows the Lewis structure and the calculations
Answers
Answers
Answer:
See explanation below
Explanation:
To solve this problem, we need to use the expression of half life decay of concentration (or mass) which is the following:
m = m₀e^-kt (1)
In this case, k will be the constant rate of this element. This is calculated using the following expression:
k = ln2/t₁/₂ (2)
Let's calculate the value of k first:
k = ln2/2.7 = 0.2567 d⁻¹
Now, we can use the expression (1) to calculate the remaining mass:
m = 8.1 * e^(-0.2567 * 2.6)
m = 8.1 * e^(-0.6674)
m = 8.1 * 0.51303
m = 4.16 mg remaining
Final answer:
The half-life of gold-198 is the time it takes for half of it to decay. Given that the half-life is 2.7 days, and the period in consideration is 2.6 days, approximately half of the original amount of 8.1 mg, which is 4.05 mg, will remain.
Explanation:
This problem is related to the concept of half-life in radioactive decay. The half-life of a substance is the time it takes for half of it to decay. As the half-life of gold-198 is 2.7 days and we are considering a period of 2.6 days, which is almost one half-life, therefore, approximately half the substance should have decayed.
So, if you start with 8.1 mg of gold-198, at the end of one half-life (or close to it at 2.6 days), you should have approximately half of this amount remaining. Half of 8.1 mg is 4.05 mg, thus, approximately 4.05 mg remains after 2.6 days.
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