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Radda [10]
3 years ago
10

What is the mass of 7.91 cm to the power of three piece of lead having a density of 11.34 g/cm to the power of three?

Chemistry
1 answer:
miss Akunina [59]3 years ago
4 0

Hey there!

D = 11.34 g/cm³

V = 7.91 cm³

Therefore:

m = D * V

m = 11.34 * 7.91

m = 89.6994 g

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timurjin [86]

B. The energy barrier between reactants and products

hope this helps!

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7 0
2 years ago
Read 2 more answers
After 20 min, a reactant has decomposed to 85 % of its original concentration. Which order would the reaction need to be to allo
rjkz [21]

Answer : The value of rate constant is, 0.0949\text{ min}^{-1}

Explanation :

First we have to calculate the rate constant, we use the formula :

Expression for rate law for first order kinetics is given by:

k=\frac{2.303}{t}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time passed by the sample  = 20 min

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process = 100 - 85 = 15 g

Now put all the given values in above equation, we get

k=\frac{2.303}{20}\log\frac{100}{15}

k=0.0949\text{ min}^{-1}

Therefore, the value of rate constant is, 0.0949\text{ min}^{-1}

7 0
3 years ago
What is Darwin's name for species that do not appear to have changed for millions of years?
ioda

Since, the options have not been given the question is incomplete.

What is Darwin's name for species that do not appear to have changed for millions of years?

a.

Dinosaurs

b.

Living fossils

c.

Old souls

d.

Ancient moderns

Answer: b. Living fossil

Explanation:

In 1859 Charles Darwin proposed the term living fossil, that means a species or group of species that had not changed in terms of evolutionary context thus can be useful in tracing the extinct or previously existing forms of life. The examples of the living fossils are horseshoe crabs, ginkgo (Conifers) and tuatara. These group of animals were existed unchanged in the Ordovician, Permian, and Triassic periods respectively with few surviving species.

4 0
2 years ago
Given the following: I) N20(g) 1/2 02(g) 2 NO(g) II) N2(g) 02(g) 2 NO(g) Ke= 4.1 x 10-31 Ke= 1.7 x 10-13 Find the value of the e
tresset_1 [31]

Answer:

Answer B) 4.2x10^17

Explanation:

To produce the reaction 3 using reaction 1 and 2 we need to invert the order of the first reaction the second in the same order, as it's shown:

2NO-------->N_2O+1/2O_2 : (reaction 1' :K'_1 =1/K_1=2.4x10^ 3^0)

N_2+O_2-------->2NO: (reaction 2 :K_2 =1.7x10^-^1^3)

____________________________

N_2+1/2O_2-------->N_2O: (reaction 3)

Due to the inversion of the first equation, the equilibrium constant of the new reaction is K1'=1/K1.=2.4x10^30

Finally, the new equilibrium constant K3 is the product of the previous constants:  

K3=K1'*K2=4.2x10^17

6 0
3 years ago
A 0.552-g sample of ascorbic acid (vitamin C) was dissolved in water to a total volume of 20.0 mL and titrated with 0.1103 M KOH
Tanzania [10]

Answer:

176.36g/mol

Explanation:

It was given that:

Mass of ascorbic acid=0.552 g

Volume of water=20.0 mL

Concentration of KOH=0.1103 M

Volume of KOH=28.42 mL. = 0.02842l

pH of solution at 10.0 mL= 3.72

At equivalence point, number of moles of acid is equal to the number of moles of base.

Number of moles of base= (KOH) x Volume

                        =0.1103 x 0.02842 L

                                =0.00313 moles.

Therefore number of miles of acid= 0.00313moles

No of moles= Mass/molar mass

Molar Mass= Mass of Acid/ No of moles of acid

= 0.552g/0.00312moles

= 176.36g/mol

4 0
3 years ago
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