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nikklg [1K]
3 years ago
5

At another temperature, the rate constant is 3.25 M-2 s-1. What is the rate at this temperature for the

Chemistry
1 answer:
brilliants [131]3 years ago
4 0

Answer:

It's 0.00015

Explanation:

got it right on edge 2021

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You are the medical examiner at the scene of a murder at a restaurant. There is a
Hoochie [10]

The compound : C₄₀H₄₄N₄O

<h3>Further explanation</h3>

The empirical formula is the smallest comparison of atoms of compound =mole ratio of the components

The principle of determining empirical formula

• Determine the mass ratio of the constituent elements of the compound.  

• Determine the mole ratio by dividing the percentage by the atomic mass

The mol ratio of composition : C : H : N : O

\tt \dfrac{80.66}{12}\div \dfrac{7.39}{1}\div \dfrac{9.39}{14}\div \dfrac{2.68}{16}\\\\6.722\div 7.39\div 0.671\div 0.1675\rightarrow divide~by~smallest(0.1675)\\\\40\div 44\div 4\div 1

3 0
3 years ago
In a solution of a carbonated beverage the water is what? Solute, saturated, solvent or precipitate
Scilla [17]
Https://us-static.z-dn.net/files/d49/33d4ec86853ef95e6f6c14242c663be4.png
5 0
3 years ago
A) 30.22 g NaCl x 1 mol NaC l =      58.4430 Molar mass (g) NaCl=0.5171 mol NaCl
jekas [21]
The third question requires you to solve for the weight of sodium (Na) and weight of Chloride (Cl) from the calculated moles of each element Na, and Cl.

So, you need to multiply the calculated moles of Na with its molar mass (23 g/ mol) to get the answer for Na. And multiply the calculated moles of Cl with its molar mass (35.45 g/mol) to get the answer for Cl. 
6 0
3 years ago
A 6.165 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 10.27 grams of CO2 and 3
e-lub [12.9K]

Answer:

Explanation:

mass of carbon in 10.27 g of CO₂ = 12 x 10.27 / 44 = 2.80 g

mass of hydrogen ( H ) in 3.363 g of H₂O = 2  x 3.363 / 18

= .373 g

These masses would have come from the sample of 6.165 g .

Rest of 6.165 g of sample is oxygen .

So oxygen in the sample = 6.165 - ( 2.8 + .373 ) = 2.992 g

Ratio of C  , H , O in the sample

2.8 : .373 : 2.992

C: H : O : : 2.8 : .373 : 2.992

Ratio of moles

C: H : O : : 2.8/12 : .373/1 : 2.992 / 16

C: H : O : : .2333 : .373 : .187

C: H : O : : .2333/.187 : .373/.187 : .187/.187

C: H : O : : 1.247 : 1.99 : 1

C: H : O : : 5 : 8 : 4 ( after multiplying by 4 )

Hence empirical formula

C₅H₈O₄

Molecular formula ( C₅H₈O₄ )n

n ( 5 x 12 + 8 x 1 + 4 x 16 ) = 132

n x ( 60 + 8 + 64 ) = 132

n = 1

Molecular formula = C₅H₈O₄.

3 0
3 years ago
What is the boiling point of water when 175.0 g of Na2SO4, a strong electrolyte is dissolved in 1.000 Kg of water?
liubo4ka [24]

Answer: 101.9^0C

Explanation:

Elevation in boiling point is given by:

\Delta T_b=i\times K_b\times m

\Delta T_b=T_b-T_b^0=(T_b-100)^0C = Elevation in boiling point

i= vant hoff factor = 3 (number of ions an electrolyte produce on complete dissociation)

Na_2SO_4\rightarrow 2Na^++SO_4^{2-}

K_f = freezing point constant = 0.512^0C/m

m= molality

\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (water)= 1.000 kg

Molar mass of solute Na_2SO_4 = 142 g/mol

Mass of solute Na_2SO_4  = 175.0 g

(T_b-100)^0C=3\times 0.512\times \frac{175.0g}{142g/mol\times 1.000kg}

T_b=101.9^0C

Thus the boiling point of water when 175.0 g of Na_2SO_4, a strong electrolyte is dissolved in 1.000 Kg of water is 101.9^0C

8 0
3 years ago
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