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3 years ago

Why is the moon far away but look so close

Chemistry

1 answer:

Roman55 [17]3 years ago

5 0

because the government is lying to us about the moon and everything in space.

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P4O10(s) = -3110 kJ/mol H2O(l) = -286 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in enthalpy for the following process:

mash [69]

Answer:

-290KJ/mol

Explanation:

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}

ΔHrxn = 4(-1279) - [6(-286) - 3110]

= -5116 -(-1716-3110)

= -5116-(-4826)

= -5116 + 4826 = -290KJ/mol

6 0

3 years ago

What is the atomic weight of a hypothetical element consisting of two isotopes, one with mass = 64.23 amu (26.0%), and one with

matrenka [14]

I think it should be A or D

8 0

3 years ago

Submit What is the solubility of Cd3(POA) 2 in water? (Ksp of Cd3(PO4)2 is 2.5 x 10-33) | 1 2 3 +/- . 0 x100

vesna_86 [32]

<u>Answer:</u> The solubility of $Cd_3(PO_4)_2$ in water is $1.18\times 10^{-7}mol/L$

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

$Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}$

3s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2$

We are given:

$K_{sp}=2.5\times 10^{-33}$

Putting values in above equation, we get:

$2.5\times 10^{-33}=(3s)^3\times (2s)^2\\\\2.5\times 10^{-33}=108s^5\\\\s=1.18\times 10^{-7}mol/L$

Hence, the solubility of $Cd_3(PO_4)_2$ in water is $1.18\times 10^{-7}mol/L$

6 0

3 years ago

Using the systematic approach for equilibrium problems, calculate the pH of 0.05 M HOCl. Ka= 3.0*10-8 Group of answer choices 3.

SVEN [57.7K]

Answer:

The pH is equal to 4.41

Explanation:

Since HClO is a weak acid, its dissociation in aqueous medium is:

HClO ⇄ ClO- + H+

start: 0.05 0 0

change -x +x +x

balance 0.05-x x x

As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.

the acidity constant when equilibrium is reached is equal to:

$Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}$

The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:

$3x10^{-8}=\frac{x^{2} }{0.05}$

clearing the x and calculating its value we have:

$x=3.87x10^{-5}=[H+]=[ClO-]$

the pH can be calculated by:

$pH=-log[H+]=-log[3.87x10^{-5}]=4.41$

7 0

3 years ago

in which part of the plant does photosynthesis take place A. the roots B. The leaves C. the flowers?

kondaur [170]

Answer:

Explanation:

3 0

3 years ago

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