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Solnce55 [7]
2 years ago
8

A) How many grams of Cu(OH)2will precipitate when excess NaOH solution is added to 42.0mL of 0.641 M CuI2 solution?

Chemistry
1 answer:
Dmitriy789 [7]2 years ago
6 0
The answer would possibly be 25g
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Which of the following lists the elements in order of increasing malleability
dimulka [17.4K]

a...if you had the fully extended periodic table, these elements would be between La and Hf in period 6 and Ac and Rf in period 7

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3 years ago
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What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon
Varvara68 [4.7K]

Answer: 31.8 g

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Al_2O_3=\frac{60.0g}{102g/mol}=0.59moles

\text{Moles of} C=\frac{30.0g}{12g/mol}=2.5moles

Al_2O_3+3C\rightarrow 2Al+3CO  

According to stoichiometry :

1 mole of Al_2O_3 require 3 moles of C

Thus 0.59 moles of Al_2O_3 will require=\frac{3}{1}\times 0.59=1.77moles  of C

Thus Al_2O_3 is the limiting reagent as it limits the formation of product and C is the excess reagent as it is present in more amount than required.

As 1 mole of Al_2O_3 give = 2 moles of Al

Thus 0.59 moles of Al_2O_3 give =\frac{2}{1}\times 0.59=1.18moles  of Al

Mass of Al=moles\times {\text {Molar mass}}=1.18moles\times 27g/mol=31.8g

Thus 31.8 g of Al will be produced from the given masses of both reactants.

5 0
3 years ago
W myFLVS
bazaltina [42]

Answer:

.. / -.. --- -. - / -.- -. --- .-- / .-- .... .- - / .. / .-- .- ... / - .... .. -. -.- .. -. --. / .-.. . .- ...- .. -. --. / -- -.-- / -.-. .... --- .. .-.. -.. / -... . .... .. -. -.. / -. --- .-- / .. / ... ..- ..-. ..-. . .-. / .- / -.-. ..- .-. ... . / .- -. -.. / .. / .- -- / -... .-.. .. -. -.. / - .... .-. ..- / .- .-.. .-.. / - .... .. ... / .- -. --. . .-. / --. ..- .-.. - / ... .- -.. -. . ... ... / -.-. --- -- .. -. --. / - --- / .... .- ..- -. - / -- . / ..-. --- .-. . ...- . .-. / .. / -.-. .- -. - / .-- .- .. - / ..-. --- .-. / - .... . / -.-. .-.. .. ..-. ..-. / .- - / - .... . / . -. -.. / --- ..-. / - .... . / .-. .. ...- . .-. .-.-.- / .. ... / - .... .. ... / .-. . ...- . -. --. . / - .... .- - / .. -- / ... . . -.- .. -. --. ..--.. / --- .-. / ... . . -.- .. -. --. / ... --- -- --- -. . / - --- / .- ...- . -. --. . / -- . ..--.. / ... - ..- -.-. -.- / .. -. / -- -.-- / --- .-- -. / .--. .- .-. .- -.. --- -..- / .. / .-- .- -. .- / ... . - / -- -.-- / ... . .-.. ..-. / ..-. .-. . . .-.-.-

Explanation:

5 0
3 years ago
A carbon forms one bond to an X atom plus three more bonds to three hydrogen atoms. If the hybridization to X has 81% p-characte
liraira [26]

Answer:

110 degree

Explanation:

This is because Hybridization of an s orbital with all three p orbitals (px , py, and pz) results in four sp3 hybrid orbitals. sp3 hybrid orbitals are oriented at bond angle of 109.5 degrees from each other. This 109.5 degrees gives an arrangement of tetrahedral geometry

5 0
3 years ago
The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures
QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

7 0
2 years ago
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