Chemistry, 02.12.2021 21:10, jakalenn2018

Subject A (without diabetes) Food type Starting blood glucose level (mg/dL) Blood glucose level after 1 hour (mg/dL) Blood glucose level after 2 hours (mg/dL) Blood glucose level after 3 hours (mg/dL) 100 100 100 100 Subject B (with type 1 diabetes) Food type Starting blood glucose level (mg/dL) Blood glucose level after 1 hour (mg/dL) Blood glucose level after 2 hours (mg/dL) Blood glucose level after 3 hours (mg/dL) 100 100 100 100

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answered: Guest

out of these options:

ability to burn paper

pure sodium reacting violently with water

the boiling point of water

helium does not tend to react with anything

i would say this is a physical property:

helium does not tend to react with anything

why? :

i said helium does not tend to react with anything because well, it reacts to ! but ability to burn paper, pure sodium reacting violently with water, & the boiling point of water are chemical or

physical properties: the properties of a substance that can be observed without changing the chemical makeup of the substance

chemical properties: properties of a substance relating to the chemical nature and reactivity of the substance

answered: Guest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

answered: Guest

For cyclohexane compounds drawn in a chair conformation: -axial bonds are vertical -equatorial bonds are parallel to two bonds within the ring. thus, as i assume the chair conformation is drawn: -the x group is in an axial position, as are the five other vertical bonds. -three axial bonds are on the top face of the ring -the other three axial bonds are on the bottom face the position of the second substituent on each ring carbon is equatorial. gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other. drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1. click this link and watch some academic videos that may you more understand the topic

answered: Guest

football

explanation:

because it is