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olga nikolaevna [1]

3 years ago

8

Which of these is a characteristic of science? (5 points) Question 1 options: 1) It cannot be reproduced by any scientist. 2) It

cannot be changed by new knowledge. 3) It is based on empirical evidence. 4) It is guided by a set of values.

1 answer:

Alex17521 [72]3 years ago

4 0

Answer:

3

Explanation:

It is based on empirical evidence

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We can consider a chemical to be safe if:

slamgirl [31]

We can consider a chemical to be safe if it does not contain any harmful substances such as nitrogen, harmful acids, or even excessive heat. Other chemicals such as the chemicals used in food products like, for example, citric acid, is not so harmful.
Hope this helps! :D
Any questions? Just let me know! I'd be happy to help any way possible.

8 0

3 years ago

A gas containing 80% CH4 and 20% He is sent through a quart/ diffusion tube (see Figure P8.8) to recover the helium, Twenty perc

elena55 [62]

For a gas containing 80% CH4 and 20% He is sent through a quart diffusion tube, the composition is mathematically given as

%He=12.5%

%CH4=87.5%

<h3>What is the composition of the waste gas if 100 kg moles of gas are processed per minute?</h3>

Generally, the equation for the Material balance is mathematically given as

F=R+W

Therefore

100=0.20*1000+W

W=80kmol/min

In conclusion, waste gas compose

2.0/100*100=50/100*20+%*80

Hence

%He=12.5%

%CH4=87.5%

Read more about Chemical reaction

brainly.com/question/16416932

3 0

2 years ago

For the reaction between aqueous solutions of acetic acid (CH3COOH) and barium hydroxide, Ba(OH)2, 1. Write the balanced molecul

Crazy boy [7]

Answer:

$2CH_3COOH(aq)+Ba(OH)_2(aq)\rightarrow Ba(CH_3COO)_2(aq)+2H_2O(l)$

Explanation:

When acetic acid solution and barium hydroxide solution react together to give an aqueous solution of barium acetate and water

The balanced chemical reaction will be given by

$2CH_3COOH(aq)+Ba(OH)_2(aq)\rightarrow Ba(CH_3COO)_2(aq)+2H_2O(l)$

8 0

3 years ago

Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

$C_p-C_v=R$

As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0

3 years ago

A 65.0-gram sample of some unknown metal at 100.0° C is added to 100.8 grams of water at 22.0° C. The temperature of the water r

vampirchik [111]

We do a heat balance to solve this:
(m cp ΔT)water = -(m cp ΔT)metal
100.8 (4.18) (27 - 22) = -65 (cp)(27-100)
cp = 100.8 (4.18) (27 - 22) / (-65 (27-100))
cp = 0.44 J/ (°C × g)

The specific heat of the metal is 0.44 J/ (°C × g)

5 0

3 years ago