Answer:
Explanation:
Hello!
In this case, we can divide the problem in two steps:
1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:
So we solve for C2:
2. Now, since 111 mL of water is added, we compute the final volume, V3:
So, the final concentration of the 139 mL portion is:
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B is the correct answer for it
Answer:
2J/g°C
Explanation:
Q = 5000J
Initial temperature (T1) = 20°C
Final temperature (T2) = 70°C
Specific heat capacity (c) = ?
Heat energy (Q) = mc∇T
Q = mc∇T
Q = mc(T2 - T1)
5000 = 50 × c × (70 - 20)
5000 = 50c × 50
5000 = 2500c
c = 5000 / 2500
c = 2J/g°C
The specific heat capacity of the substance is 2J/g°C
<span>c. About one month
To answer this question, TAKE A LOOK AT THE GRAPH. If you do so, you'll see that the first peak for prey happens at about 2.5 months. The first peak for predators happens at about 3.5 months, or in other words, lags by about a month. Looking at the second peak for prey and predator you see the figures of 8 months and 9 months. Another lag of about 1 month. Looking at the third peak, you see a bit past 13 months and a bit past 14 months. Another one month lag. Therefore the answer is "c. About one month"</span>
Answer:
0.5mol/L
Explanation:
First, let us calculate the number of mole NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the question = 30g
Number of mole = Mass /Molar Mass
Number of mole = 30/40 = 0.75mol
Volume = 1.5L
Active mass = mole/Volume
Active mass = 0.75mol/1.5L
Active mass = 0.5mol/L
