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faltersainse [42]

3 years ago

5

How is data not actually obtained from the experiment represented in a line graph? with a double line with only dots with a colo

red line with a broken line

2 answers:

alexandr1967 [171]3 years ago

8 0

Answer: The answer is broken line

Fittoniya [83]3 years ago

7 0

The way how <span>data is not actually obtained from the experiment represented in a line graph is defnitely that </span><span>a colored line with a broken line. It is a well known fact that to obtain the actual data from the experiment you there should be plotted points on the line. Hope it will help you! Regards.</span>

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How many milligrams of sodium sulfide are needed to completely react with 25.00 ml of a 0.0100 m aqueous solution of cadmium nit

NARA [144]

Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)

v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol

n(Na₂S)=n{Cd(NO₃)₂}=cv

m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv

m(Na₂S)=78.046*0.0100*25.00≈19.5 mg

5 0

4 years ago

Read 2 more answers

A cylinder with a moveable piston contains 92g of Nitrogen. The external pressure is constant at 1.00 atm. The initial temperatu

Jobisdone [24]

Answer:

Work done in this process = 4053 J

Explanation:

Mass of the gas = 0.092 kg

Pressure is constant = 1 atm = 101325 pa

Initial temperature $T_{1}$ = 200 K

Final temperature $T_{2}$ = 200 - 85 = 115 K

Gas constant for nitrogen = 297 $\frac{J}{kg k}$

When pressure of a gas is constant, volume of the gas is directly proportional to its temperature.

⇒ V ∝ T

⇒ $\frac{V_{2} }{V_{1} }$ = $\frac{T_{2} }{T_{1} }$ ------------ ( 1 )

From ideal gas equation $P_{1}$ $V_{1}$ = m R $T_{1}$ ------ (2)

⇒ 101325 × $V_{1}$ = 0.092 × 297 × 200

⇒ $V_{1}$ = 0.054 $m^{3}$

This is the volume at initial condition.

From equation 1

⇒ $\frac{V_{2} }{0.054}$ = $\frac{200}{115}$

⇒ $V_{2}$ = 0.094 $m^{3}$

This is the volume at final condition.

Thus the work done is given by W = P [ $V_{2}$ - $V_{1}$ ]

⇒ W = 101325 × [ 0.094 - 0.054]

⇒ W = 4053 J

This is the work done in that process.

7 0

3 years ago

What problems could arise if scientists use different units to measure height

Shkiper50 [21]

Scientists can measure the height in different units but problem could arise when they compare all the measurements. That is the reason there is standard units for measurements.

<span>There may be error arises when an American scientist is measuring the height of an object in inches and other Australian scientist is measuring the height of same object in meters. Their data cannot be compared because they are using different units to measure height.</span>

5 0

3 years ago

(35 points) ROCK TYPES. If you help me with this I'll love you forever

4vir4ik [10]

There are so many rock types around! Though, with all the rock types.. there are a few that we study within grade school and such. These types of rocks are,

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4 0

4 years ago

14. 60. g of NaOH is dissolved in enough distilled water to make 300 mL of a stock solution. What volumes of this solution and d

zepelin [54]

The question is incomplete, the complete question is attached below.

Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

Explanation : Given,

Mass of NaOH = 60 g

Volume of stock solution = 300 mL

Molar mass of NaOH = 40 g/mol

First we have to calculate the molarity of stock solution.

$\text{Molarity}=\frac{\text{Mass of }NaOH\times 1000}{\text{Molar mass of }NaOH\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M$

Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock solution.

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution.

From data (A) :

$M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 20mL=1M\times V_2\\\\V_2=100mL$

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (B) :

$M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 20mL=1M\times V_2\\\\V_2=100mL$

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

$M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 60mL=1M\times V_2\\\\V_2=300mL$

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

$M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 60mL=1M\times V_2\\\\V_2=300mL$

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

5 0

3 years ago