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Andre45 [30]
3 years ago
12

Calculate ΔG ° for the following reactions at 25 ° C:

Chemistry
1 answer:
Maksim231197 [3]3 years ago
7 0

Answer:

NOPE

Explanation:

DAT IS TO HARD! WATCH SOME TIKTOK!

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a sample of 25.0g of an unknown metal is added to 25.0ml of water in a graduated cylinder and the final volume is 28.5ml what is
leva [86]

Answer:

<h2>The answer is 7.14 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass of metal = 25 g

volume = final volume of water - initial volume of water

volume = 28.5 - 25 = 3.5 mL

It's density is

density =  \frac{25}{3.5}  \\  = 7.142857...

We have the final answer as

<h3>7.14 g/mL</h3>

Hope this helps you

8 0
3 years ago
I know that one of them is multiple choice but i’m not sure which one it is.
Evgen [1.6K]

Explanation:

1. neutrons

2. protons

3. protons and neutrons

4. neutrons

5. electrons

I'm not sure with 1 and 4, sorry :<

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3 years ago
Common effects of cocaine include
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4 years ago
A gas mixture with a total pressure of 765 mmHg contains each of the following gases at the indicated partial pressures: 134 mmH
ladessa [460]

Answer: c

Explanation:

4 0
3 years ago
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0
3 years ago
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