State whether each of the following equations represents a synthesis (s), decomposition(d), single replacement

ipn [44]

Answer:

isolated system (plural isolated systems) (physics) A system that does not interact with its surroundings. Depending on context this may mean that its total energy and/or momentum stay constant.

Explanation:

An isolated system is a thermodynamic system that cannot exchange either energy or matter outside the boundaries of the system. ... The system may be enclosed such that neither energy nor mass may enter or exit.

is there both?

7 0

3 years ago

Calculate the mass of liquid with density of 3.3 g/mL and a volume of 25 oz

love history [14]

Answer: 215

Explanation:

Volume doesn’t add up into ounces that’s mass only

3 0

3 years ago

Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

3 years ago

PLS help

viktelen [127]

Answer:

d . Sc2O5

Explanation:

Hello,

In this case, when forming oxides from a metal and oxygen, for us to find out each element's subscript, we must exchange them as shown below, considering +5 for scandium:

$Sc^{+5}_2O^{-2}_5$

For that reason, the answer is d . Sc2O5

Best regards.

3 0

3 years ago

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Strontium sulfate becomes less soluble in an aqueous solution when sodium sulfator is added because

horsena [70]

Answer:

The addition of sulfate ions shifts equilibrium to the left.

Explanation:

Hello!

In this case, according to the following ionization of strontium sulfate:

$SrSO_3(s)\rightleftharpoons Sr^{2+}+SO_4^{2-}$

It is evidenced that when sodium sulfate is added, sulfate, $SO_4^{2+}$ is actually added in to the solution, which causes the equilibrium to shift leftwards according to the Le Ch athelier's principle. Thus, the answer in this case would be:

The addition of sulfate ions shifts equilibrium to the left.

Best regards!

7 0

2 years ago

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