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natima [27]
2 years ago
8

What are the atomic and mass numbers for the periodic table of elements

Chemistry
1 answer:
chubhunter [2.5K]2 years ago
8 0

there you go, hope that helps

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Óxido cuprico+amoniaco-------Nitrógeno +cobre +agua(redox)
tigry1 [53]

Answer:

?? no entiendo la pregunta?...pero gracias por los puntos free UvU

Explanation:

3 0
2 years ago
The teacher said the volume of the liquid was 500 mL when measured a student found it was 499.7 mL what is the students percent
Natalka [10]

Answer:

<h2>0.06 % </h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

error = 500 - 499.7 = 0.3

actual volume = 500 mL

We have

p(\%) =  \frac{0.3}{500}  \times 100 \\  =  \frac{3}{50}  \\

We have the final answer as

<h3>0.06 % </h3>

Hope this helps you

4 0
3 years ago
When an acid reacts with a base, salt and water is formed! Why??​
vladimir1956 [14]

Answer:

When n acid reacts with a base , salt and water is formed because when acid reacts with base, it loses its acidic property and base loses its basic property to form a neutral substance like salt and water..

5 0
3 years ago
Read 2 more answers
Vitamin K is involved in normal blood clotting. When 2.09 g of vitamin K is dissolved in 25.0 g of camphor, the freezing point o
AURORKA [14]
I don't think there is a question in there
7 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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