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3 years ago

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7th grade science can anyone help

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Assoli18 [71]3 years ago

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Who was given name black hole

marin [14]

Answer:

John Archibald Wheeler

3 0

3 years ago

How many moles of electrons are transferred when 2.0 moles of aluminum metal react with excess copper(II) nitrate in aqueous sol

mojhsa [17]

Moles of electrons:

The moles of electrons that are transferred are 12F

A balanced equation:

2 moles of Aluminium metal react with excess copper(II) nitrate.

$2Al + 3Cu{(NO_3)}_2 \rightarrow 2Al{(NO_3)}_3 +3 Cu$

Given:

Moles of Aluminium = 2

As Aluminium goes from 0 to +3 oxidation state

$Al \rightarrow Al^{3+} + 3e^-$

And copper goes from +2 to 0

$Cu^{2+} + 2e^-\rightarrow Cu$

On balancing the number of electrons we get:

For 1 mole of Al $6e^-$ is required.

Therefore for 2 moles of Al,

Total $(2\times6)$ F mole of electrons

Where F= Faraday's constant= 96500 C

So, 12F moles of electrons are transferred.

Learn more about Faraday's Law here,

brainly.com/question/27985929

#SPJ4

4 0

1 year ago

The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i

Rus_ich [418]

Answer: a) The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

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

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

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) concentration after 8.8 min:

$8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}$

$\log\frac{0.25}{a-x}=0.15$

$\frac{0.25}{a-x}=1.41$

$(a-x)=0.17M$

b) for concentration to decrease from 0.25M to 0.15M

$t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20$

$t=687s$

7 0

3 years ago

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago

Fill in the blank.

labwork [276]

Answer:

Kinetic energy, Kelvin temperature, direction, warmer, cooler, Kelvin, Absolute zero

Explanation:

THe answers are in order and when there is a comma move to the next ______

7 0

2 years ago