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Arte-miy333 [17]

3 years ago

5

In an experiment, you measure the density of aluminium as 2.60 g/cm^3 . The accepted value is 2.70 g/cm^3 . What is the percent

of error in your measurement?

1 answer:

FinnZ [79.3K]3 years ago

4 0

5.30% im pretty sure of it

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Which correctly lists the types of matter, in terms of their atoms, in order from least tightly packed to most tightly packed?

SSSSS [86.1K]

Answer:

D

Explanation:

gas are well separated with no regular arrangement and spreaded all over. liquid are close together with no regular arrangement and slightly spreaded. solid are tightly packed, usually in a regular pattern and not spreaded at all.

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3 years ago

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Valence electrons are: O a. Electrons that have been lost in an atom O b. Electrons in the nucleus of an atom O c. electrons in

mestny [16]

C ) outer most shell of atom

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2 years ago

What quantity of copper is deposited by the same quantity of electricity that deposited 9g of aluminum

klio [65]

Answer:

Mass of copper deposited = 31.75 g

Explanation:

According to Faraday's second law of electrolysis, when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements deposited are inversely proportional to the charges on the ions of the elements.

From this law, it can be seen that the higher the charge, the lower the number of moles of a given element deposited.

Number of moles of aluminium in 9 g of aluminium = mass / molar mass

Molar mass of aluminium = 27 g

Number of moles of aluminium = 9/27 = 1/3 moles

Charge on aluminium ion = +3

3 moles of electrons will discharge 1 mole of aluminium,

1 mole of electrons will discharge 1/3 moles of aluminium

Number of moles of electrons involved = 1 mole of electrons

Charge on copper ion = +2

1 mole of electrons will discharge 1/2 moles of copper.

Mass of 1/2 moles of copper = number of moles × molar mass of copper

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Mass of copper deposited = 1/2 × 63.5 = 31.75 g

3 0

3 years ago

Combustion of methanol displayed formula???

Andrew [12]

Answer:

reee, here is your answer.

Explanation: CH3OH(l) + 3O2(g) rightarrow CO2(g) + 3H2O(g) CH3OH(l) + O2(g) rightarrow CO2(g) + 2H20(g) CH3OH(l) + 2O2(g) rightarrow 2CO2(g) + 4H20(g) 2CH3OH(l) + 3O2(g) rightarrow 2CO2(g) + 4H20(g) Correct Calculate Delta H degree rxn at 25 degree C.

6 0

3 years ago

A sealed 1.0 L flask is charged with 0.500 mol of I2 and 0.500 mol of Br2. An equilibrium reaction ensues: I2 (g) + Br2 (g) ↔ 2I

Daniel [21]

Answer: Thus the value of $K_{eq}$ is 110.25

Explanation:

Initial moles of $I_2$ = 0.500 mole

Initial moles of $Br_2$ = 0.500 mole

Volume of container = 1 L

Initial concentration of $I_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M$

Initial concentration of $Br_2=\frac{moles}{volume}=\frac{0.500moles}{1L}=0.500M$

equilibrium concentration of $IBr=\frac{moles}{volume}=\frac{0.84mole}{1L}=0.84M$ [/tex]

The given balanced equilibrium reaction is,

$I_2(g)+Br_2(g)\rightleftharpoons 2IBr(g)$

Initial conc. 0.500 M 0.500 M 0 M

At eqm. conc. (0.500-x) M (0.500-x) M (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[IBr]^2}{[Br_2]\times [I_2]}$

$K_c=\frac{(2x)^2}{(0.500-x)\times (0.500-x)}$

we are given : 2x = 0.84 M

x= 0.42

Now put all the given values in this expression, we get :

$K_c=\frac{(0.84)^2}{(0.500-0.42)\times (0.500-0.42)}$

$K_c=110.25$

Thus the value of the equilibrium constant is 110.25

8 0

3 years ago