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musickatia [10]
2 years ago
12

Can you put this in order from least to greatest for me {335 B.C. 430 B.C. 400 B.C.}

Chemistry
2 answers:
AlekseyPX2 years ago
8 0
It would be 430 B.C. 400 B.C. 335 B.C.
I hope this helps ;)
notka56 [123]2 years ago
8 0
First it's 335 B.C. , next it's 400 B.C. , then lastly it's 430 B.C.  

The same way you put this in order from "least to greatest" ... It's just like Mathematics terms of "least to greatest" .. Just with 'B.C.' added to it 
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How many kj of heat are needed to completely melt 32.3 g of h2o, given that the water is at its melting point? the heat of fusio
timurjin [86]

Answer: fourth option, 10.8 kJ


Explanation:


The <em>heat of fusion</em>, also named latent heat of fusion, is the amount of heat energy required to change the state of a substance from solid to liquid (at constant pressure).


The data of the <em>heat of fusions</em> of the substances are reported in tables and they can be shown either per mole or per gram of substance.


In this case we have that the<em> heat of fusion for water </em>is reported per mole: <em>6.02 kJ/mole</em>.


The formula to calculate <em>how many kJ of heat (total heat) are needed to completely melt 32.3 g of water, given that the water is at its melting point</em> is:

  • Heat = number of moles × heat of fusion

The calculations are:

  • number of moles = mass / molar mass

        number of moles = 32.3 g / 18.015 g/mol = 1.79 mol

       

  • Heat = 1.79 mol × 6.02 kJ / mol = 10.8 kJ ← answer
5 0
3 years ago
Read 2 more answers
A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h
frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]

Where;

\frac{f_2}{f_1}  is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1}  = 10^{0.7289}\\\\\frac{f_2}{f_1}  = 5.356

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0
3 years ago
A 2.5 L sample of gases at STP (standard temperature and pressure is 273 k and 1.00 atm). When the temperature is raised to 273d
podryga [215]

Answer:

x = 5.9

Explanation:

6 0
3 years ago
Chlorine has two stable isotopes, Cl-35 and Cl-37. If their exact masses are 34.9689 amu and 36.9695 amu, respectively, what is
natima [27]

Answer:

X(Cl-35) = 75.95% => Answer 'A'

Explanation:

34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance

X(Cl-35) + X(Cl-37) = 1  ⇒  X(Cl-37) = 1 - X(Cl-25)

34.9689·X(Cl-35) + 36.9695(1 - X(Cl-35)) = 35.45

34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45

Rearrange ...

36.9695·X(Cl-35) - 34.9689·X(Cl-35) = 36.9689 - 35.45

2.0006·X(Cl-35) = 1.5195

X(Cl-35) = 1.5195/2.0006 = 0.7595 fractional abundance

⇒ % abundance = 75.95%

3 0
3 years ago
Read 2 more answers
1. The multiplicative inverse of 5/9 is.....<br>a-9/5<br>b-9/5<br>c-3/9<br>d-None of these​
Luda [366]

Answer:

None of these cause the correct answer is 9/-5

5 0
3 years ago
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