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kicyunya [14]
3 years ago
11

Which choice lists the correct order of the coefficients of each substance in the following neutralization reaction when the equ

ation is balanced correctly? H2SO4 + LiOH → Li2SO4 + H2O 1, 1, 1, 1 2, 1, 2, 1 1, 2, 1, 2 1, 2, 1, 1
Chemistry
2 answers:
inessss [21]3 years ago
3 0
I got a 4 out of 1. not 100 percent sure if this is the right answer but i put 1,2,1,2
shtirl [24]3 years ago
3 0

Answer : The order of the coefficients of each substance in the following neutralization reaction are, 1, 2, 1, 2

Explanation :

Balanced chemical reaction : It is defined as the chemical reaction in which the number of individual atoms of an element in reactant side always be equal to the number of individual atoms of an element in product side.

The given unbalanced reaction will be,

H_2SO_4+LiOH\rightarrow Li_2SO_4+H_2O

In order to balance the chemical reaction, the coefficient 2 is put before the reactant LiOH and product H_2O.

The balanced chemical reaction will be,

H_2SO_4+2LiOH\rightarrow Li_2SO_4+2H_2O

By the stoichiometry we can say that, 1 moles of H_2SO_4 react with 2 moles of LiOH to give 1 mole of Li_2SO_4 and 2 moles of H_2O as a product.

Hence, the order of the coefficients of each substance in the following neutralization reaction are, 1, 2, 1, 2

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The pressure of the gas used in the weather balloon increases to expand the balloon.

Explanation:

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8 0
4 years ago
Can some one please help me with this organic chemistry question?
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I think it is gauche.

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3 years ago
A dull metal object has a density of 8.8 G/ML and a volume of 20 ML calculate the mass
Alex

Answer:

Mass = 0.000176 gram

Steps:

m =  V × ρ

=  20 milliliter × 8.8 gram/cubic meter

=  2.0E-5 cubic meter × 8.8 gram/cubic meter

=  0.000176 gram

Explanation:

8 0
3 years ago
What is 5.0553 x 10^-4 in standard notation?
Talja [164]

Answer:

0.00050553

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6 0
3 years ago
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
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