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LUCKY_DIMON [66]

2 years ago

12

A solution has a higher boiling point than its associated pure solvent does.

1 answer:

marta [7]2 years ago

7 0

Answer:

4 boiling point elevation

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Muscle contraction is triggeredA) in response to an increase in the cytoplasmic Ca2+concentration.B) in response to a decrease i

natima [27]

Answer:

A) in response to an increase in the cytoplasmic Ca2+concentration.

Explanation:

Muscle contraction occurs in response to an increase in the cytoplasmic Ca2 + concentration.

This process occurs with the shortening of the sarcomeres resulting in a result, the actin filaments react with myosin, generating actomyosin. During this reaction, it is necessary to increase the cytoplasmic concentration of Ca + and ATP. In this, myosin will break down ATP, releasing energy so that the muscle can contract.

4 0

2 years ago

What is the value of ΔG in a spontaneous reaction?

Amanda [17]

The value of Triangle G is A) Less than 0.

7 0

2 years ago

Read 2 more answers

Which of the following are balanced equations? Check all that apply.

EleoNora [17]

Answer: the answer is b

Explanation:

because why not

8 0

2 years ago

Caluclate the number of grams of oxygen that must react

bonufazy [111]

Answer : The number of grams of oxygen react must be, 170.4 grams.

Solution : Given,

Mass of $C_3H_8$ = 46.85 g

Molar mass of $C_3H_8$ = 44 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $C_3H_8$ .

$\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{46.85g}{44g/mole}=1.065moles$

Now we have to calculate the moles of $O_2$

The balanced chemical reaction is,

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

From the balanced reaction we conclude that

As, 1 mole of $C_3H_8$ react with 5 mole of $O_2$

So, 1.065 moles of $C_3H_8$ react with $1.065\times 5=5.325$ moles of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(5.325moles)\times (32g/mole)=170.4g$

Therefore, the number of grams of oxygen react must be, 170.4 grams.

8 0

2 years ago

3.0 mL of 0.02 M Fe(NO3)3 solution is mixed with 3.0 mL of 0.002 M NaNCS and diluted to the mark with HNO3 in 10 mL volumetric f

Stella [2.4K]

Answer:

The $K_c$ for $[Fe(NCS)]^{2+}$ formation is $5.7\times 10^2$ .

Explanation:

$Moles=Concentration\times Volume (L)$

$Fe(NO_3)_3(aq)\rightarrow Fe^{3+}(aq)+3NO_3^{-}(aq)$

$[Fe(NO_3)_3]=0.02 M=[Fe^{3+}]$

Concentration of ferric ion = $[Fe^{3+}]=0.02 M$

Volume of ferric solution = 3.0 mL = 0.003 L

Moles of ferric ion $=0.02 M\times 0.003 L$

1 mL = 0.001 L

$NaNCS(aq)\rightarrow Na^+(aq)+NCS^-(aq)$

$[NaNCS]=0.002 M=[NCS^-]$

Concentration of $NCS^-$ ion = $[NCS^{-}]=0.002 M$

Volume of $NCS^-$ ion solution = 3.0 mL = 0.003 L

Moles of $NCS^-$ ion= $0.002 M\times 0.003 L$

Volume of nitric acid solution = 10 mL = 0.010 L

After mixing all the solution the concentration of ferric ion and $NCS^-$ ion will change

Total volume of solution = 0.003 L + 0.003 L + 0.010 L = 0.016 L

Initial concentration of ferric ion before reaching equilibrium :

= $\frac{0.02 M\times 0.003 L}{0.016 L}=0.00375 M$

Initial concentration of $NCS^-$ ion before reaching equilibrium :

= $\frac{0.002 M\times 0.003 L}{0.016 L}=0.000375 M$

$Fe^{3+}+NCS^-\rightleftharpoons [Fe(NCS)]^{2+}$

Initially:

0.00375 M 0.000375 M 0

At equilibrium :

(0.00375-x) (0.000375-x) x

Equilibrium concentration of $[Fe(NCS)]^{2+}=x=2.5\times 10^{-4} M$

The expression of equilibrium constant for formation $[Fe(NCS)]^{2+}$ is given by :

$K_c=\frac{[[Fe(NCS)]^{2+}]}{[Fe^{3+}][NCS^-]}$

$K_c=\frac{x}{(0.00375-x)\times (0.000375-x)}$

$K_c=\frac{2.5\times 10^{-4} }{(0.00375-2.5\times 10^{-4})\times (0.000375-2.5\times 10^{-4})}$

$K_c=5.7\times 10^2$

The $K_c$ for $[Fe(NCS)]^{2+}$ formation is $5.7\times 10^2$ .

5 0

3 years ago