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riadik2000 [5.3K]

3 years ago

14

A student performs an acid-base titration experiment to determine the amount of ascorbic acid in different brands of juice. Whic

h solution should the student use as the base solution in this acid-base titration experiment?
A.) HCl
B.) NaOH
C.) LiBr
D.)CaCl2

1 answer:

Mice21 [21]3 years ago

8 0

Answer:

idn Mann nnnnnnnn . .....mmmmm mmmm aaaaaaa it's a

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What is the correct equation for the solubility product when calcium phosphate is dissolved in water

AlladinOne [14]

Answer:

The solubility of calcium phosphate is

2.21x10−4g/L.

Explanation:

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3 years ago

Elements whose names end with âium are usually metals; sodium is one example. Identify a nonmetal whose name also ends with âium

Makovka662 [10]

Answer: Helium (He) and Selenium (Se) both are non metals ends with ium sound

Explanation:

He is a inert gas and Se belongs to oxygen family (VI A group)

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3 years ago

16. A sample of nitrogen gas,

Lana71 [14]

find mol of N2 present using gas law equation

PV = nRT

P = pressure = 688/760 = 0.905 atm.

V = 100mL = 0.1L

n = ???

R = 0.082057

T = 565+273 = 838

Substitute:

0.905*0.1 = n*0.082057*838

n = 0.0905 / 68.76

n = 0.00132 mol N2

Molar mass N2 = 28 g/mol

0.00132 mol = 0.00132*28 = 0.037g N2 gas

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3 years ago

Which property of metals allows aluminum to be flattened into thin sheets of aluminum foil?

Gnoma [55]

Answer:

malleability

Explanation:

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3 years ago

Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C

umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

$K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}$

Given:

$\left[I_2_{Equilibrium} \right]$ = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

$0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}$

So,

$\left[I_{Equilibrium} \right]^2=0.011\times 0.10$

$\left[I_{Equilibrium} \right]^2=0.0011$

$\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M$

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0

4 years ago