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slega [8]
3 years ago
11

I + I -----> I2 + 146.3 kJ Is this reaction exothermic or endothermic?

Chemistry
1 answer:
elixir [45]3 years ago
6 0
This reaction is endothermic!
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Does a negative exponent mean that the number is less than 1 yes or no​
Masja [62]

Answer:

When a number is written in scientific notation, the exponent tells you if the term is a large or a small number. A positive exponent indicates a large number and a negative exponent indicates a small number that is between 0 and 1.

3 0
3 years ago
I don't know how to do this, pls help!
Aliun [14]
Cations from smallest to largest
Li⁺ ,Na⁺, K⁺ (from Periodic Table, the bigger number of period, the bigger size, of atom, so the bigger size of cation)

1) LiF smaller  cation then KF
1,036                                 <span>853
</span><span>The lattice energy increases as cations get smaller, as shown by LiF and KF.
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2) The same cation Li , so wrong statement.

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4) NaCl smaller cation then KF
  786                                   853
7 0
3 years ago
Read 2 more answers
An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr
Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa

Rate of flow of ideal gas , n = 4 kmol/hr = \frac{4\times 1000mol}{3600s}=1.11mol/s    (Conversion factors used:  1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW     (Conversion factor:  1 kW = 1000 W)

We know that:

\Delta U=0   (For isothermal process)

So, by applying first law of thermodynamics:

\Delta U=\Delta q-\Delta W

\Delta q=\Delta W      .......(1)

Now, calculating the work done for isothermal process, we use the equation:

\Delta W=nRT\ln (\frac{P_1}{P_2})

where,

\Delta W = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

P_1 = initial pressure = 100 kPa

P_2 = final pressure = 50 kPa

Putting values in above equation, we get:

\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0
3 years ago
How many electrons does the copper atom lose in the process: Cu → Cu3+ + ? electrons
alexdok [17]

Answer:

3

Explanation:

6 0
3 years ago
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