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3 years ago

In a chemical equation, the sum of the masses of the products is equal to the sum of the masses of the reactants. True False

Chemistry

2 answers:

Komok [63]3 years ago

8 0

The answer is true e e e e e e e e e e

Elan Coil [88]3 years ago

8 0

Your answer should be true

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One of these element behaves partly as the alkaline metal as well as a halogen element

goldenfox [79]

The answer is Hydrogen

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3 years ago

The boiling process is considered to be____

Dmitry_Shevchenko [17]

your answer is D. physical change, because a new substance is not formed.

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2 years ago

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vess

Tpy6a [65]

Answer:

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

Explanation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The relation of $K_c\& K_p$ is given by:

$K_p=K_c(RT)^{\Delta n_g}$

$K_p$ = Equilibrium constant in terms of partial pressure.=98.1

$K_c$ = Equilibrium constant in terms of concentration =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side= $n_{g,p}-n_{g.r}=1-2=-1$

$98.1=K_c(RT)^{-1}$

$98.1 =\frac{K_c}{RT}$

$K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}$

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

5 0

3 years ago

Read 2 more answers

How would granite be classified as

vredina [299]

Answer:

Granite is an intrusive igneous rock which means it is cooled slowly deep upper the Earth's crust. It is composed of 25% to 35% quartz and over 50% potassium- and sodium rich feldspars.

Explanation:

4 0

2 years ago

Read 2 more answers

How would you prepare 3.5 L of a 0.9M solution of KCl?

Fudgin [204]

Calculate the mass of the solute in the solution :

Molar mass KCl = 74.55 g/mol

m = Molarity * molar mass * volume

m = 0.9 * 74.55 * 3.5

m = 234.8325 g

To prepare 0.9 M KCl solution, weigh 234.8325 g of salt in an analytical balance, dissolve in a beaker, shortly after transfer with the help of a funnel of transfer to a volumetric flask of 100 cm³ and complete with water up to the mark, then cover the balloon and finally shake the solution to mix

hope this helps!

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3 years ago