Answer : The rate constant at 785.0 K is, 
Explanation :
According to the Arrhenius equation,

or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at
= 
= rate constant at
= ?
= activation energy for the reaction = 262 kJ/mole = 262000 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature = 
Now put all the given values in this formula, we get:
![\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B6.1%5Ctimes%2010%5E%7B-8%7Ds%5E%7B-1%7D%7D%29%3D%5Cfrac%7B262000J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B600.0K%7D-%5Cfrac%7B1%7D%7B785.0K%7D%5D)

Therefore, the rate constant at 785.0 K is, 
he required empirical formula based on the data provided is Na2CO3.H2O.
<h3>What is empirical formula?</h3>
The term empirical formula refers to the formula of a compound which shows the ratio of each specie present.
We have the following;
Mass of sodium = 37.07-g
Mass of carbonate = 48.39 g
Mass of water = 14.54-g
Number of moles of sodium = 37.07-g/23 g/mol = 2 moles
Number of moles of carbonate = 48.39 g/61 g/mol = 1 mole
Number of moles of water = 14.54/18 g/mol = 1 mole
The mole ratio is 2 : 1: 1
Hence, the required empirical formula is Na2CO3.H2O
Learn more about empirical formula : brainly.com/question/11588623
Answer:
Halophile.
Explanation:
Halophile microorganisms are microorganisms that require very large amounts of NaCl. If the concentration of NaCl is very little, there will be no growth. Ih this case, given that it grows between 5% and 15%, it can be considered a moderate halophile.
D all are 1 because the left side equals the right
Answer:
The mass of copper(II) sulfide formed is:
= 81.24 g
Explanation:
The Balanced chemical equation for this reaction is :

given mass= 54 g
Molar mass of Cu = 63.55 g/mol

Moles of Cu = 0.8497 mol
Given mass = 42 g
Molar mass of S = 32.06 g/mol

Moles of S = 1.31 mol
Limiting Reagent :<em> The reagent which is present in less amount and consumed in a reactio</em>n
<u><em>First find the limiting reagent :</em></u>

1 mol of Cu require = 1 mol of S
0.8497 mol of Cu should require = 1 x 0.8497 mol
= 0.8497 mol of S
S present in the reaction Medium = 1.31 mol
S Required = 0.8497 mol
S is present in excess and <u>Cu is limiting reagent</u>
<u>All Cu is consumed in the reaction</u>
Amount Cu will decide the amount of CuS formed

1 mole of Cu gives = 1 mole of Copper sulfide
0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide
= 0.8497
Molar mass of CuS = 95.611 g/mol


Mass of CuS = 0.8497 x 95.611
= 81.24 g