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3 years ago

8

Lakes, rivers, and streams are created due to what stage of the water cycle?

2 answers:

a_sh-v [17]3 years ago

6 0

When molecules of water vapor return to liquid or solid form they create like cloud droplets that fall back into earth as rain or snow that process is Condensation. which that’s the precipitation that falls into rivers, streams, etc

VMariaS [17]3 years ago

3 0

Answer:

r

Explanation:

r

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What is the hydrogen ion concentration of 0.450M solution of acetic acid (HC2H3O2), if the ionization constant(Ka) is 2.19 x 10-

WITCHER [35]

Answer:

3.02 Mm

Explanation:

3 0

3 years ago

What is the energy of a photon of green light with a frequency of 5.50 x 101^4 Hz?

Minchanka [31]

I didn't know if you meant to the power of 14 but if you did here your answer:
3.64 x 10^-19
(you just multiply the frequency by Planck's constant= 6.63 × 10^–34)

8 0

3 years ago

For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

7 0

3 years ago

What text features were used? font color, shadow, and bold font color, italics, and outline font color, italics, and bold font c

AlladinOne [14]

Answer:

Explanation:

question isnt clear

4 0

3 years ago

When 33 g of CaO and 10 g of H2O react, how many grams of calcium hydroxide would you expect to be produced* Explain your answer

MrMuchimi

Answer:

41.44 g

Explanation:

First of all, we must put down the equation of the reaction;

$CaO + H2O ----->Ca(OH)2$

Number of moles of CaO = 33g/56 g/mol = 0.59 moles

Number of moles of H20 = 10g/18 g/mol = 0.56 moles

Since the reaction is in 1:1 mole ration, H2O is the limiting reactant

Hence;

mass of Ca(OH)2 produced = 0.56 moles * 74 g/mol = 41.44 g

7 0

3 years ago

Read 2 more answers