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3 years ago

Which substance can not be decomposed by a chemical change?

Chemistry

2 answers:

oksian1 [2.3K]3 years ago

8 0

The substance that can not be decompoased by a chemical change is B. Carbon

FrozenT [24]3 years ago

3 0

Carbon because it an element in comparison to eater for example which is a mixture of hydrogen and oxygen.

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Three separate 3.5g blocks of al, cu, and fe at 25°c each absorb 0.505 kj of heat. which block reaches the highest temperature?

77julia77 [94]

The energy can be shown as:

Q = ms dT

Whereas, m is the mass of block

s is specific heat

dT is change in temperature.

Copper block having the lowest specific heat and thus having the higher change in temperature and therefore having the higher final temperature.

8 0

3 years ago

What does phenolphthalein turn pink?

Alex Ar [27]

Phenolphthalein is often used as an indicator in acid–base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions. Phenolphthalein is slightly soluble in water and usually is dissolved in alcohols for use in experiments.

8 0

3 years ago

Convert 3 moles of h2o into grams

lilavasa [31]

To convert 3 mol H2O to grams, just multiply by the molar mass of H2O.

<span>3 mol H2O * 18 g H2O / 1 mol H2O = 54 g H2O

Hope this helped.</span>

4 0

3 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago

Determine the equilibrium constant, Kp, for the following reaction, by using the two reference equations below: 2 NO(g) + O2(g)

klio [65]

Answer:

$Kp=3.07x10^6$

Explanation:

Hello,

In this case, by knowing the given reference reactions, one could rearrange them as follows:

$2 NO(g) \leftrightarrow N_2(g) + O_2(g); Kp_2 = \frac{1}{2.3 x 10^{-19}}=4.35x10^{18}$

$N_2(g) + 2O_2(g) \leftrightarrow 2NO_2(g);Kp_3=(8.4x10^{-7})^2=7.056x10^{-13}$

Subsequently, to obtain the main reaction, we add the aforementioned reference rearranged reactions as shown below (just as reference):

$2NO(g)+N_2(g)+2O_2\leftrightarrow 2NO_2(g)+N_2+O_2$

Consequently, the equilibrium constant is computed as:

$Kp=\frac{[N_2][O_2]}{[NO]^2} * \frac{[NO_2]^2}{[N_2][O_2]^2} =Kp_2*Kp_3=4.35x10^{18}*7.056x10^{-13}=3.07x10^6$

Best regards.

8 0

3 years ago