Answer:
a ice cube on a room temperature rock
Explanation:
because the air temperature is warmer than the freezers which means the ice particle gather heat energy from warmer air.
The balanced nuclear equations for the following:(a) β⁻ decay of silicon-32 is (27,14)Si -> (0,-1)beta + (27,15)P
<h3>
What is balanced nuclear equation?</h3>
A nuclear reaction is generally expressed by a nuclear equation, which has the general form, where T is the target nucleus, B is the bombarding particle, R is the residual product nucleus, and E is the ejected particle, and Ai and Zi (where I = 1, 2, 3, 4) are the mass number and atomic number, respectively. Finding a well balanced equation is critical for understanding nuclear reactions. Balanced nuclear equations provide excellent information about the energy released in nuclear reactions. Balancing the nuclear equation requires equating the total atomic number as well as the total mass number before and after the reaction using the rules of atomic number and mass number conservation in a nuclear reaction.
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Answer:
Explanation:
We will need a balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
Mᵣ: 44.01
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
n/mol: 1.5
1. Calculate the moles of CO₂
The molar ratio is 3 mol CO₂:1 mol C₃H₈
2. Calculate the mass of CO₂.
MThe heat energy required to raise the temperature of 0.36Kg of copper from 22 c to 60 c is calculate using the following formula
MC delta T
m(mass)= 0.360kg in grams = 0.360 x1000 = 360 g
c(specific heat energy) = 0.0920 cal/g.c
delta T = 60- 23 = 37 c
heat energy is therefore= 360g x0.0920 cal/g.c x 37 c= 1225.44 cal
Answer:
![[Cu^{2+}]=0.041 M](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%20M)
Explanation:
Hello!
In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:
![[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D%3D0.041%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D%2A%5Cfrac%7B1molCu%5E%7B2%2B%7D%7D%7B1molCu%28NH_3%29_4%5E%7B2%2B%7D%7D%20%3D0.041%20%5Cfrac%7BmolCu%28NH_3%29_4%5E%7B2%2B%7D%7D%7BL%7D)
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