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3 years ago

11

which piece of equipment would be used to measure the volume of a liquid? a. a beaker b. a meter stick c. a graduated cylinder d

. a triple-beam balance

1 answer:

Ulleksa [173]3 years ago

3 0

It is a graduated cylinder.

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Given the unbalanced equation:

maw [93]

Answer:

Option (2) 2

Explanation:

NO3- + 4H+ + Pb → Pb2+ + NO2 + 2H2O

The equation above can be balance as follow:

There are 3 atoms of the left side and a total of 4 atoms on the right side. It can be balance by putting 2 in front NO3- and 2 in front of NO2 as shown below:

2NO3- + 4H+ + Pb → Pb2+ + 2NO2 + 2H2O

Now the equation is balanced.

The coefficient of NO2 is 2

8 0

3 years ago

A graduated cylinder contains 20.5 mL of water. What is the new water level after 35.2 g of silver metal with a density of 10.5

iragen [17]

<span>V = 24.0 mL + (35.2 g)(mL/10.5g) = I think i'm not all that sure but I think its this.</span>

3 0

3 years ago

Read 2 more answers

Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress

ElenaW [278]

Answer:

(a) $W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) $W_{polytropic}=7.579\frac{kJ}{mol}$

(c) $W_{isothermal}=5.743\frac{kJ}{mol}$

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant $k$ should be computed for air as an ideal gas by:

$\frac{R}{Cp_{air}}=1-\frac{1}{k} \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\$

$0.2856=1-\frac{1}{k}\\\\k=1.4$

Next, we compute the final temperature:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K$

Thus, the work is computed by:

$W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}$

(b) In this case, since $n$ is given, we compute the final temperature as well:

$T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K$

And the isentropic work:

$W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}$

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

$W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}$

Regards.

8 0

3 years ago

Please help and thank you so much

olchik [2.2K]

A. Sorry if I’m wrong :(

8 0

3 years ago

What did the scientists who founded the royal scientists society of london share with lavoisier

dmitriy555 [2]

<span>they both had their conclusions based on solid evidence</span>

6 0

3 years ago