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3 years ago

Please help with all 3 parts!

Chemistry

2 answers:

Murljashka [212]3 years ago

8 0

Answer:

1:Part A.

$\bold{42.2 g C_{12}H_{22}O_{11} \:in \:528 g H₂O}$

Mass Percent= $\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}$

= $\frac{42.2}{528}*100\%=\bold{\underline{7.99\: or \:8\%}}$

Part B.

$\bold{198\:m g\: C_{6}H_{12}O_{6} \:in\:4.71 g\: H₂O}$

mass of solute: 198mg

mass of solvent :4.71g=4710g

Mass Percent= $\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}$

= $\frac{198}{4710}*100\%=\bold{\underline{4.20\%}}$

Part C.

$\bold{8.85 g NaCl \:in \:190 g\: H₂O}$

Mass Percent= $\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}$

= $\frac{8.85}{190}*100\%=\bold{\underline{4.66\%}}$

Alecsey [184]3 years ago

6 0

Answer:

It will help you !!!!!!!!!!

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2 years ago

For a reaction at equilibrium, which change can increase the rates of the forward and reverse reactions?a decrease in the concen

Monica [59]

Question requires a change resulting in an increase in both forward and reverse reactions. Now lets discuss options one by one and see there impact on rate of reactions.

1) <span>A decrease in the concentration of the reactants:
When concentration of reactant is decreased it will shift the equilibrium in Backward direction, so resulting in increasing the backward reaction and decreasing the forward direction. Hence, this option is incorrect.

2) </span><span>A decrease in the surface area of the products:
Greater the surface Area greater is the chances of collision and greater will be the rate of reaction. As the surface area of products is decreased it will not favor the backward reaction. Hence again this statement is incorrect according to given statement.

3) </span><span>An increase in the temperature of the system:
An increase in temperature will shift the reaction in endothermic side. Hence, if the reaction is endothermic, an increase in temperature will increase the rate of forward direction or if the reaction is exothermic it will increase the rate of reverse direction. Hence, this option is correct according to given statement.

4) </span><span>An increase in the activation energy of the forward reaction:
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3 years ago

Write a balanced equation for the complete oxidation reaction that occurs when ethane burns in air

damaskus [11]

<h3>Balanced equation : 2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)</h3><h3>Further explanation</h3>

Alkanes are saturated hydrocarbons that have single bonds in chains

General formula for alkanes :

$\tt \large{\bold{C_nH_{2n+2}}$

Hydrocarbon combustion reactions (specifically alkanes)

$\large {\box {\bold{C_nH _ (_2_n _ + _ 2_) + \dfrac {3n + 1} {2} O_2 \Rightarrow nCO_2 + (n + 1) H_2O}}}$

So that the burning of ethane with air (oxygen):

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or we can use mathematical equations to solve equilibrium chemical equations by giving the coefficients for each compound involved in the reaction

C₂H₆ (g) + aO₂ (g) ⟶ bCO₂ (g) + cH₂O (ℓ)

C : left 2, right b ⇒ b=2

H: left 6, right 2c⇒ 2c=6⇒ c= 3

O : left 2a, right 2b+c⇒ 2a=2b+c⇒2a=2.2+3⇒2a=7⇒a=7/2

6 0

3 years ago

The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that

kakasveta [241]

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is $1.1\times 10^6$ M.

$HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)$

This value indicates that

A. $CN^-$ is a stronger base than $ONO^-$

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is $ONO^-$

D. The conjugate acid of CN- is HCN

Answer: A. $CN^-$ is a stronger base than $ONO^-$

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When $K_{p}>1$ ; the reaction is product favoured.

When $K_{p};$ ; the reaction is reactant favored.

$When K_{p}=1$ ; the reaction is in equilibrium.

As, $K_p>>1$ , the reaction will be product favoured and as it is a acid base reaction where $HONO$ acts as acid by donating $H^+$ ions and $CN^-$ acts as base by accepting $H^+$

Thus $HONO$ is a strong acid thus $ONO^-$ will be a weak conjugate base and $CN^-$ is a strong base which has weak $HCN$ conjugate acid.

Thus the high value of K indicates that $CN^-$ is a stronger base than $ONO^-$

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3 years ago