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3 years ago

HELP ME ASAP !!!!!!!!!!

Chemistry

2 answers:

Ugo [173]3 years ago

4 0

Answer:

running 20 m/s south

Explanation:

so its c

Cloud [144]3 years ago

3 0

Answer:

the yellow one

Explanation:

it includes direction

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What is the wavelength of a photon with an energy of 3.50 x 10^-19 J ?

Leokris [45]

Answer:

λ = 5.68×10⁻⁷ m

Explanation:

Given data:

Energy of photon = 3.50 ×10⁻¹⁹ J

Wavelength of photon = ?

Solution:

E = hc/λ

h = planck's constant = 6.63×10⁻³⁴ Js

c = 3×10⁸ m/s

Now we will put the values in formula.

3.50 ×10⁻¹⁹ J = 6.63×10⁻³⁴ Js × 3×10⁸ m/s/ λ

λ = 6.63×10⁻³⁴ Js × 3×10⁸ m/s / 3.50 ×10⁻¹⁹ J

λ = 19.89×10⁻²⁶ J.m / 3.50 ×10⁻¹⁹ J

λ = 5.68×10⁻⁷ m

8 0

3 years ago

what is the concentration of hydroxide ions after 50.0 ml of 0.250 m naoh is added to 120 ml of 0.200 m na2so4? please show all

Galina-37 [17]

The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

What is meant by concentration?

Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.

Concentration of hydroxide ions can be calculated by,

M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.

where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.

Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.

To learn more about concentration click on the given link brainly.com/question/17206790

#SPJ4

5 0

1 year ago

1-chloro-4-methyl-2-pentene undergoes hydrolysis in warm water to give a mixture of 4-methyl-2-penten-1-ol and 4-methyl-1-penten

agasfer [191]

Answer:

Explanation:

attached here is the diagram representing the structure

4 0

3 years ago

a. Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g

Tanya [424]

Answer:

(A) $\Delta H^{\circ }_{r}= -144 kJ$

(B) $\Delta H^{\circ }_{r}= - 2552kJ$

Explanation:

(A) 2NO(g) + O₂(g) → 2NO₂(g)

$1/2 N_{2}(g)+O_{2}(g)\rightarrow NO_{2}(g), \Delta H^{\circ }_{a}=33.2 kJ....equation (a)$

$1/2N_{2}(g)+1/2O_{2}(g)\rightarrow NO(g), \Delta H^{\circ }_{b}=90.2 kJ ....equation (b)$

Now, multiplying equation (a) with 2:

⇒ $N_{2}(g)+2 O_{2}(g)\rightarrow 2 NO_{2}(g)....equation (a)$

Then equation b is reversed and multiplied with 2:

$2 NO(g)\rightarrow N_{2}(g)+ O_{2}(g)....equation (b)$

Now by adding the equation (a) and equation (b), we get:

⇒ $2 NO(g)+ \bcancel N_{2}(g)+\bcancel 2 O_{2}(g)\rightarrow 2 NO_{2}(g) +\bcancel N_{2}(g)+ \bcancel O_{2}(g)$

⇒ 2NO(g) + O₂(g) → 2NO₂(g)

<u>Therefore, the enthalpy of the reaction:</u>

$\Delta H^{\circ }_{r}= 2\times \Delta H^{\circ }_{a} - 2\times \Delta H^{\circ }_{b}$

$= (2\times33.2)- (2\times90.2)=66.4 - 180.4= -144 kJ$

(B) 4B(s)+3O₂(g) → 2B₂O₃(s)

$B_{2}O_{3}(s)+3H_{2}O(g)\rightarrow 3O_{2}(g)+B_{2}H_{6}(g), \Delta H_{a }^{\circ }=+2035 kJ...equation (a)$

$2B(s)+3H_{2}(g)\rightarrow B_{2}H_{6}(g), \Delta H_{b }^{\circ }= +36 kJ...equation (b)$

$H_{2}(g)+1/2O_{2}(g)\rightarrow H_{2}O(l), \Delta H_{c }^{\circ }= -285 kJ...equation (c)$

$H_{2}O(l)\rightarrow H_{2}O(g), \Delta H_{d }^{\circ }=+44 kJ...equation (d)$

Now multiplying equation (b) with 2, reversing equation (a) and multiplying with 2. Reversing equation (c) and (d) and multiplying both with 6.

$6O_{2}(g)+2B_{2}H_{6}(g)\rightarrow 2B_{2}O_{3}(s)+6H_{2}O(g)...equation (a)$

$4B(s)+6H_{2}(g)\rightarrow 2B_{2}H_{6}(g)...equation (b)$

$6H_{2}O(l)\rightarrow 6H_{2}(g)+3O_{2}(g)...equation (c)$

$6H_{2}O(g)\rightarrow 6H_{2}O(l)...equation (d)$

Now by adding the equations (a), (b), (c), (d); we get:

4B(s)+3O₂(g) → 2B₂O₃(s)

<u>Therefore, the enthalpy of the reaction: </u>

$\Delta H^{\circ }_{r}= -2\times \Delta H^{\circ }_{a} + 2\times \Delta H^{\circ }_{b} - 6 \times \Delta H_{c }^{\circ } - 6 \times \Delta H_{d }^{\circ }$

$= -2\times (+2035 kJ)+ 2\times (+36 kJ) - 6 \times (-285 kJ)- 6 \times (+44 kJ) = -4070 + 72 + 1710 - 264 = - 2552kJ$

4 0

3 years ago

For the reaction: 4PH3(g) → P4(g) + 6H2(g) about 0.065 mol/s of PH3 is consumed in a 5.0 L flask. What are the rates of producti

hoa [83]

Answer: The rates of production of $P_4$ is $3.25\times 10^{-3}$ mol/Ls and $H_2$ is 0.0195 mol/Ls.

Explanation:

$4PH_3(g)\rightarrow P_4(g)+6H_2(g)$

Rate with respect to reactants is shown by negative sign as the reactants are decreasing with time and Rate with respect to products is shown by positive sign as the products are increasing with time.

Rate of the reaction= $-\frac{1}{4}\frac{[d[PH_3]}{dt}=\frac{[d[P_4]}{dt}=\frac{1}{6}\frac{[d[H_2]}{dt}$

Rate of decomposition of $PH_3=\frac{0.065 mol/s}{5.0L}=0.013mol/Ls$

Rate of production of $P_4=\frac{1}{4}\times {\text{rate of decomposition of}} PH_3=\frac{1}{4}\times 0.013=3.25\times 10^{-3}mol/Ls$

Rate of production of $H_2=\frac{6}{4}\times {\text {rate of decomposition of}} PH_3=\frac{6}{4}\times 0.013=0.0195mol/Ls$

7 0

3 years ago

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