Can someone please help me with this? <br>

What is the enthalpy change, ΔH, for this reaction? Show your work.

zubka84 [21]

Answer:

B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.

Explanation:

B) What is the enthalpy change, ∆H, for this reaction? Show your work to receive full credit (5 points) The enthalpy change is 150. To find it we must subtract energy of products (200) & the energy of reactants (50) so 200 – 50 equals 150.

3 0

3 years ago

Read 2 more answers

Lance Armstrong bikes at a constant speed up the Col d’Izoard, a famous mountain pass. Assume his teammates do such a good job r

svetoff [14.1K]

Work done against gravity to climb upwards is always stored in the form of gravitational potential energy

so we can say

$W = mgh$

here h = vertical height raised

so here we know that

$h = 14.1 sin7.3 km$

here we have

$h = 1.79 km$

now from above equation

$W = (83 kg)(9.81 m/s^2)(1.79 \times 10^3 m)$

$W = 1.46 \times 10^6 J$

so work done will be given by above value

7 0

3 years ago

Why are there multiple versions of the scientific method?

gavmur [86]

Answer:

it's because some versions have more steps and others have less

4 0

3 years ago

An ant travels 2.78 cm (West) and then turns and travels 6.25 cm (South 40 degrees East). What is the ant's total displacement?

andrey2020 [161]

$From\ cosine\ theorem:\\\\
c^2=a^2+b^2-2abcos(\angle between\ a\ and\ b)\\\\
a=2,78\\b=6,25\\
\angle between\ a\ and\ b=90+50=140^{\circ}\\\\
c^2=2,78^2+6,25^2-2*2,78*6,25cos(140^{\circ})\\\\
c^2=7,7284+39,0625-34,75*(-0,77)\\\\c^2=46,7909+26,7575\\\\c^2=735484\\\\c=8,58cm\\\\Total\ displacement\ is\ equal\ to\ 8,58cm.$

5 0

3 years ago

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago

Can someone please help me with this? ​

Can someone please help me with this?