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3 years ago

All of the following are possible sources of error in a scientific investigation except for

2 answers:

6 0

Answer: Common sources of error include instrumental, environmental, procedural, and human. All of these errors can be either random or systematic depending on how they affect the results. Make sure you have your problem, hypothesis, evidence, analyze the data, ask yourself if the evidence supports ur hypothesis, draw conclusions, and communicate your results!

Explanation:

3241004551 [841]3 years ago

3 0

Eviormental bc with out the environment to test on scientists wouldn’t be able to do anything bc they would t be alive

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How consumers affect which goods and services are produced.

oksano4ka [1.4K]

Consumers affect the goods and services produced through:

- The more we buy something the more popular it is and the more of it is made

- The products that is less popular and less made are because we don't buy them

- When we buy more of something the more expensive the product or service will be

- When we buy less of something the more cheaper that product or service will be

Please vote my answer branliest! Thanks.

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3 years ago

If a gas is initially at a pressure of nine ATM and a volume at 21 L at a temperature of 253K and the pressure is raise to 15 AT

Tom [10]

Answer:

15.0L

Explanation:

p/v = constan

(9*21)/253 =(15v)/ 302

v = (9*21*302)/(15*253)

v=15.0

3 0

3 years ago

Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C

umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

$K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}$

Given:

$\left[I_2_{Equilibrium} \right]$ = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

$0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}$

So,

$\left[I_{Equilibrium} \right]^2=0.011\times 0.10$

$\left[I_{Equilibrium} \right]^2=0.0011$

$\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M$

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0

3 years ago

What are these smooth, distinct layers most directly evidence of ?

Neko [114]

Answer: The layers of the rocks in one region of the parks are smooth and distinct, which are evidence of many, many years of deposition. The layers on the rocks are because of different deposition of sediments. Different sediments deposited over the rocks through wind, water and ice over the ages

Explanation:

5 0

2 years ago

Read 2 more answers

At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha

Alekssandra [29.7K]

Explanation:

$Density =\frac{Mass}[Volume}$

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball = $V=560 cm^3=560 ml=0.560 L$

$D =\frac{0.12 g}{0.560 L}=0.214 g/L$

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = $d_1=1.830 g/L$

Mass of the carbon dioxide gas :

$1.830 g/L\times 0.560 L=1.0248 g$

Total density of filled ball with carbon dioxide gas:

$\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L$

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = $d_2=0.0899 g/L$

Mass of the hydrogen gas :

$1.830 g/L\times 0.560 L=0.050344 g$

Total density of filled ball with hydrogen gas:

$\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L$

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = $d_3=1.330 g/L$

Mass of the oxygen gas :

$1.330 g/L\times 0.560 L=1.7448 g$

Total density of filled ball with oxygen gas:

$\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L$

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = $d_4=1.165 g/L$

Mass of the nitrogen gas :

$1.165 g/L\times 0.560 L=0.6524 g$

Total density of filled ball with nitrogen gas:

$\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L$

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

$\frac{0.12g +m}{0.560L}>1.189 g/L$

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0

2 years ago