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3 years ago

Calculate the atomic mass of nitrogen if the two common isotopes of nitrogen have masses

1 answer:

6 0

(99.63/100 x 14.003) + (0.37/100 x 15)

13.9511889 + 0.0555 = 14.0066889amu

13.95 + 0.0555 = 14.0055 (same thing just rounded numbers)

(You can round that to 14amu)

Answer is 14 amu

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PLEASE HELP I'LL MARK BRAINLIEST

dem82 [27]

Answer:

When you are in a cool place and exhale into your mask, the mask blocks and re-directs some of your breath up toward the top edge of the mask and under your glasses. ... When the warm breath hits the cool lens, the water in the breath condenses causing your lenses to fog up.

4 0

3 years ago

If an endothermic reaction begins at 26 degrees celsius and decreases by 2 degrees Celsius per minute, how long would it take to

Natalija [7]

Answer:

It would take 13 minutes.

Explanation:

The temperature decreases at a linear rate, meaning that we can describe the process by using the following formula:

T₂ = T₀ - 2*t

Where T₂ is the final temperature, T₀ is the initial temperature; and t is the elapsed time in minutes.

We input the data given by the problem:

0 °C = 26 °C - 2°C/min * t

And solve for t:

t = 13 min

8 0

3 years ago

Data for the decomposition of hydrogen peroxide at some set temperature T is provided below. The rate law depends only on the co

Viktor [21]

Answer:

$0.00442 s^{-1}$ is the value of the rate constant.

Explanation:

$2 H_2O_2\rightarrow 2 H_2O + O_2$

Let the order of the reaction be x.

The rate law of the reaction can be written as:

$R=k[H_2O_2]^x$

1. Rate of the reaction when concentration changes from 0.882 M to 0.697 M in 0 seconds to 60 seconds.

$R=-\frac{0.697 M-0.882 M}{60 s-0 s}=0.00308 M/s$

$0.00308 M/s=k[0.697 M]^x$ ..[1]

2. Rate of the reaction when concentration changes from 0.697 M to 0.566 M in 240 seconds to 360 seconds.

$R=-\frac{0.236M-0.372M}{120s-60 s}=0.00227 M/s$

$0.00218 M/s=k[0.236 M]^x$ ..[2]

[1] ÷ [2]

$\frac{0.00308 M/s}{0.00227 M/s}=\frac{k[0.697 M]^x}{k[0.236M]^x}$

Solving fro x:

x = 0.92 ≈ 1

$R=k[H_2O_2]^1$

$0.00308 M/s=k[0.697 M]^1$

$k=\frac{0.00308 M/s}{[0.697 M]^1}=0.00442 s^{-1}$

$0.00442 s^{-1}$ is the value of the rate constant.

5 0

3 years ago

A sample of a gas is occupying a 1500 ml container at a pressure of 3.4 atm and a temperature of 25 oc. if the temperature is in

Salsk061 [2.6K]

The new pressure will be 7.65 atm

<h3>General gas law</h3>

The problem is solved using the general gas equation:

P1V1/T1 = P2V2/T2

In this case, P1 = 3.4 atm, V1 = 1500 mL, T1 = 25 $^O C$ , V2 = 2000 mL, and T2 = 75 $^O C$

What we are looking for is P2.

Thus, P2 = P1V1T2/T1V2

= 3.4 x 1500 x 75/25 x 2000 = 382500/50000 = 7.65 atm

More on general gas laws can be found here: brainly.com/question/2542293

#SPJ1

7 0

2 years ago

What volume, in Liters, of a 2.00 M KCl solution contains 2.5kg of KCl? Molar mass of KCl= 74.55g/mole

Julli [10]

$C_{m}=\frac{n}{V_{r}}\\\\
n=\frac{m_{s}}{M}\\\\
C_{m}=\frac{m_{s}}{MV_{r}} \ \ \ |*MV_{r}\\\\
C_{m}MV_{r}=m_{s} \ \ \ |:C_{m}M\\\\
V_{r}=\frac{m_{s}}{C_{m}M}\\\\\\
m=2.5kg=2500g\\
M=74.55\frac{g}{mol}\\
C_{m}=2\frac{mol}{L}\\\\
V_{r}=\frac{2500g}{2\frac{mol}{L}*74.55\frac{g}{mol}}\approx16.77L$

7 0

3 years ago