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Rashid [163]
3 years ago
9

Which of these can be classified as a substance A) oxygen B) carbon dioxide C) air D) acid rain

Chemistry
2 answers:
ratelena [41]3 years ago
4 0

D could be a answer

Julli [10]3 years ago
4 0

Answer:

carbon dioxide

Explanation:

the rest are mixtures

A chemical substance is composed of one type of atom or molecule.

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Explanation:

The crabs cannot see the plankton they eat near the ocean floor. For the crabs to see the plankton, some color of visible light would need to reach the plankton so that it can be reflected into the crabs' eyes.

4 0
3 years ago
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For an ideal gas, classify the pairs of properties as directly or inversely proportional. You are currently in a sorting module.
vredina [299]

Answer:

the result for the following are (a) P is directly proportional to n

(b) V is directly proportional to T (c) P is directly proportional to T (d) T is inversly proportional to V (e) P is inversely proportional to V

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What happens to the ions when the drain is turned off? please help :] ❤️
kkurt [141]

Answer:

In a long channel MOSFET, the width of the pinch-off region is assumed small relative to the length of the channel. Thus, neither the length nor the voltage across the inversion layer change beyond the pinch-off, resulting in a drain current independent of drain bias. Consequently, the drain current saturates.

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5 0
3 years ago
A total of 20.0 mL of sodium hydroxide (NaOH) was neutralized by 30.0 mL of 0.250 M hydrogen bromide (HBr). What was the concent
Karolina [17]

Answer:

0.375 M

Explanation:

NaOH(aq) + HBr(aq) ------------> NaBr(aq) + H2O(l)

Concetration of acid CA= 0.250M

Concentration of base CB= ????

Volume of acid VA= 30.0mL

Volume of base VB= 20.0mL

Number of moles of acid nA= 1

Number of moles of base nB= 1

CA VA/CB VB= nA/nB

CB= CAVAnB/VB nA

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8 0
3 years ago
Fill in the missing data point. Show all calculations leading to an answer.
Aleksandr-060686 [28]

Answer:  

1090 mmHg  

Explanation:  

We know that with gases we must use a Kelvin temperatures, so let’s try a plot of pressure against the Kelvin temperature.  

We can create a table as follows  

<u>t/°C</u>  <u>T/K</u>  <u>p/mmHg</u>  

  10   283      726  

  20  293      750  

  40   313      800  

  70  343      880  

100  373       960  

150  423        ???  

I plotted the data and got the graph in the figure below.  

It appears that pressure is a linear function of the Kelvin temperature.  

y = mx + b  

where x is the slope and b is the y-intercept.

===============

<em>Calculate the slope  </em>

I will use the points (275, 700) and (380, 975).  

Slope = Δy/Δx = (y₂ - y₁)/(x₂ -x₁) = (975 -700)/(380 – 275) = 275/105 = 2.619  

So,  

y = 2.619x + b  

===============

<em>Calculate the intercept </em>

When x = 275, y = 700.  

700 = 2.619 × 275 + b  

700 = 720 + b     Subtract 720 from each side and transpose.  

b = -20  

So, the equation of the graph is  

y = 2.619x -20  

===============

<em>Calculate the pressure</em> at 423 K (150°C)  

y = 2.619 × 423 - 20  

y = 1110 - 20  

y = 1090

At 150 °C, the pressure 1090 mmHg.  

The point is approximately at the position of the black dot in the graph.  

7 0
3 years ago
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