What are the advantages of using an ATM? ​

1. What are the advantages of using an ATM? ​

Answer:

Atm

Atm is the most easiest to get your money

Explanation:

You can get money in it but you need an atm card

2. Using your ATM card -Advantage-Disanvantage​

Advantage:

•ATM points are conveniently located at multiple locations. You can go the ATM of any bank to withdraw cash – provided your ATM card is linked with that bank.

•There is no need to fill out withdrawal and deposit slips – unlike the practice at the bank branch.

Disadvantage:

•Fraud. Criminals can fit skimming devices and small cameras to ATMs

•Theft Risk. If you go to a bank, you're likely walking into a secured area watched by multiple cameras or a life guard. ATM doesn't have a guard.

3. what do you think are the disadvantages of opening an ATM account?​

Answer:

Hope it helps #Carryonlearning

4. At 460C a sample of ammonia gas exerts a pressure of 5.3 atm. What is the pressure when the volume of the gas is reduced to one-tenth (0.10) of the original value at the same temperature? a.53 atm b.71 atm c.45 atm d. 67 atm

Answer:

a. 53 ATM po thanks me later if correct

5. P1=1.01 atmV1=700mlP2=2.35 atmV2=​

Answer:

V1 P1 = V2 P2

(700 mL)(1.01 atm) = (V2) (2.35 atm)

1 atm = 760 torr

P1 = 1.01 atm or 767.6 torr

P2 = 2.35 atm or 1,786 torr

[tex]V2 = \frac{(700ml)(767.6mmHg)}{(1786mmHg)} [/tex]

[tex]V2 = \frac{537320ml•mmHg}{1786mmHg} [/tex]

[tex]V2 = 300.85ml[/tex]

#CarryOnLearning

#BrainlyFast

6. L tank. What is the pressure of the compressed gas, if the temperature remains constant? a. 0.15 atm b. 4.7 atm c. 0.21 atm d. 3.4 atm

3.4 atm because the neutral formula of L is given

7. A 10-liter sample of gas is held in a container under a pressure of 1.5 atm. Then gas is then compressed to 7.5 liters. Find the pressure if the temperature is kept unchanged.​A. 1.3 atmB. 2.0 atmC. 1.7 atmD. 50 atm​

Answer:

B. 2.0 atm

Explanation:

1.5 * 10 = P * 75

8. A mixture of N2 and H2 gases to be used for ammonia production exerts a pressure of 2.5 atm. If the mole fraction of N2 in the mixture is 0.67, what is the pressure exerted by N2 gas in the mixture? A. 1.70 atm B. 0.80 atm C. 1.0 atm D. 2.5 atm

Answer:

A

Explanation:

calculate it by the n2 first

9. Which of the following is used as standard pressure in a gas equation? a. 1 atm b. 2 atm c. 3 atm d. 4 atm

Answer:

1 atm is the correct answer

10. At 460C a sample of ammonia gas exerts a pressure of 5.3 atm. What is the pressure when the volume of the gas is reduced to one-tenth (0.10) of the original value at the same temperature? a.53 atm b.71 atm c.45 atm d.67 atm

Answer:

what is the different types of words and cups and cups and mixer accessories for the different options for you and your sentence ☺️ be able and willing and able to do that you something

11. 800. mL of helium is at 55.0ºC and 0.55 atm. What will the pressure be if the temperature changes to 100.ºC if the volume remains the same? a. 1 atm b. 0.62 atm c. 0.48 atm d. 0.75

The pressure if the temperature changes to 100 ºC  : 1 atm

Further explanation

Gay-Lussac's Law states that the pressure of a gas proportional to its absolute temperature of the gas when the volume is kept constant

The equation can be written :

[tex]\rm \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}[/tex]

800. mL of helium is at 55.0ºC and 0.55 atm

temperature changes to 100.ºC

So :

V (constant) = 800 ml

T1 =  55.0ºC

P1 = 0.55 atm

T2 = 100.ºC

Then :

[tex]\rm P_2=\dfrac{P_1\times T_2}{T_1}\\\\P_2=\dfrac{0.55\times 100}{55}\\\\P_2=\boxed{\bold{1~atm}}[/tex]

Learn more  

Example of ideal gas law in real life  

brainly.ph/question/2117053  

Example Problem And solution Of combined gas laws  

brainly.ph/question/2143722  

applications involving combined gas law  

brainly.ph/question/2341366  

#LetsStudy  

12. Each of the following containers has the same size, ehich of the following containers has the most compressed gas molecules? 1. 10 atm 2. 8 atm 3. 6 atm 4. 4 atm

its number one since the pressure is greatest that is 10atm so it presses the gaseous particles more than the other values given

13. A confined quantity of gas has a pressure of 2.50 atm and a temperature of -22°C. What is the new pressure if the temperature is increased to 22°C?a. 2.1 atm b. 2.5 atmc. 2.9 atmd. 3.4 atm

ANSWER: C. 2.94 L

According to Charles' Law:
[tex] \frac{pressure1}{temperature1} = \frac{pressure2}{temperature2} [/tex]
Plugging in the terms:
[tex] \frac{2.50}{251} = \frac{x}{295} \\ \frac{2.50(295)}{251} = 2.94[/tex]
The new pressure is 2.9 atm

Hello!

A confined quantity of gas has a pressure of 2.50 atm and a temperature of -22°C. What is the new pressure if the temperature is increased to 22°C ?

a. 2.1 atm  

b. 2.5 atm

c. 2.9 atm

d. 3.4 atm

We have an isochoric (isometric or isovolumetric) transformation, it is when a certain mass under pressure maintains its volume constant, if we increase the temperature increases the pressure and if we lower the temperature, the pressure also decreases.  

We have the following data:  

P1 (initial pressure) = 2.50 atm

T1 (starting temperature) = - 22 ºC  (in Kelvin)

TK = TºC + 273.15 → TK = - 22 + 273.15 → TK = 251.15

T1 (starting temperature) = 251.15 K

T2 (final temperature) = 22 ºC (in Kelvin)

TK = TºC + 273.15 → TK = 22 + 273.15 → TK = 295.15

T2 (final temperature) = 295.15 K

P2 (final pressure) = ? (atm)  

We apply the data to the formula of isovolumetric transformation (the Second Law of Charles and Gay-Lussac), let us see:

[tex]\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}[/tex]

[tex]\dfrac{2.50}{251.15} = \dfrac{P_2}{295.15}[/tex]

multiply the means by the extremes

[tex]251.15*P_2 = 2.50*295.15[/tex]

[tex]251.15\:P_2 = 73787.5[/tex]

[tex]P_2 = \dfrac{737.875}{251.15}[/tex]

[tex]P_2 = 2.93798... \to \boxed{\boxed{P_2 \approx 2.9\:atm}}\end{array}}\qquad\checkmark[/tex]

Answer:

c. 2.9 atm

______________________________

I Hope this helps, greetings ... Dexteright02! =)

14. Five gases combined in a gas cylinder have the following partial pressures: 3.00 atm (N2), 1.80 atm (O2), 0.29 atm (Ar), 0.18 atm (He), and 0.10 atm (H). What is the total pressure that is exerted by the gases?

3.00atm   +  1.80atm  +  0.29atm  + 0.18atm  +  0.10atm   =  5.37 atm

15. Three gases, Ar, N2 and H2 are mixed in a 500L container. Ar has a pressure of 2 atm, N2 has a pressure of 3 atm and H2 has pressure of 2 atm. What is the total pressure in the container?a. 2 atm b. 3 atm c. 5 atm d. 7 atm​.

Answer:

[tex] \therefore [/tex] The total pressure in the container is 7 atm (D.)

Explanation:

We have three gases which is [tex] \rm{Ar, \ N_2, \ and \ H_2} [/tex] they all mixed in 500L. So, we want to there total pressure. To know how many total of pressure in the container, we will use Dalton's Equation.

[tex] \boxed{\bold{P_{Total} = P_1 + P_2 + P_3...}}[/tex]

[tex] \begin{gathered}\begin{gathered}\begin{aligned} & \bold{Thus, \ the \ given \ are:} \\ & \quad \boxed{\begin{array}{l} \Longrightarrow \rm{P_1} = 2 \ atm \\ \Longrightarrow \rm{P_2} = 3 \ atm \\ \Longrightarrow \rm{P_3} = 2 \ atm \end{array}}\, \\ \end{aligned} \end{gathered}\end{gathered} [/tex]

Substitute it to the formula

[tex] \rm{P_{Total} = 2 \ atm + 3 \ atm + 2 \ atm} [/tex][tex] \rm{P_{Total} = 4 \ atm + 3 \ atm} [/tex][tex] \rm{P_{Total} = 7 \ atm}[/tex]

[tex] \therefore [/tex] The total pressure in the container is 7 atm (D.)

16. An ideal gas exerts a pressure of 3 atm in a 3 L container. The container is at a temperature of 298 K. What will be the final pressure if the volume of the container changes to 2 L?a. 2 atmb. 3.5 atmc. 4 atmd. 5 atmwith solution po sana! ​

Answer:

B. 3.5

Kung di ka sure o kailangan mo ng solution

download mo Yung app na Gauthmath

17. A gas mixture consists of 320. mg of methane, 175 mg of argon, and 255 mg of neon. the total pressure of the mixture is 28.3 atm. calculate the partial pressure of each gas in the mixture. [ atm, atm, atm]

SOLUTION:

Step 1: Calculate the number of moles of each gas.

The molar masses of CH₄, Ar, and Ne are 16.043 g/mol, 39.95 g/mol, and 20.18 g/mol, respectively.

One gram (g) is equal to 1000 milligram (mg).

[tex]\begin{aligned} & n_{\text{CH}_4} = \text{0.320 g} \times \frac{\text{1 mol}}{\text{16.043 g}} = \text{0.019946 mol} \\ & n_{\text{Ar}} = \text{0.175 g} \times \frac{\text{1 mol}}{\text{39.95 g}} = \text{0.0043805 mol} \\ & n_{\text{Ne}} = \text{0.255 g} \times \frac{\text{1 mol}}{\text{20.18 g}} = \text{0.012636 mol}\end{aligned}[/tex]

Step 2: Calculate the mole fraction of each gas.

• For CH₄

[tex]\begin{aligned} X_{\text{CH}_4} & = \frac{n_{\text{CH}_4}}{n_{\text{CH}_4} + n_{\text{Ar}} + n_{\text{Ne}}} \\ & = \frac{\text{0.019946 mol}}{\text{0.019946 mol + 0.0043805 mol + 0.012636 mol}} \\ & = 0.53963 \end{aligned}[/tex]

• For Ar

[tex]\begin{aligned} X_{\text{Ar}} & = \frac{n_{\text{Ar}}}{n_{\text{CH}_4} + n_{\text{Ar}} + n_{\text{Ne}}} \\ & = \frac{\text{0.0043805 mol}}{\text{0.019946 mol + 0.0043805 mol + 0.012636 mol}} \\ & = 0.11851 \end{aligned}[/tex]

• For Ne

[tex]\begin{aligned} X_{\text{Ne}} & = \frac{n_{\text{Ne}}}{n_{\text{CH}_4} + n_{\text{Ar}} + n_{\text{Ne}}} \\ & = \frac{\text{0.012636 mol}}{\text{0.019946 mol + 0.0043805 mol + 0.012636 mol}} \\ & = 0.34186 \end{aligned}[/tex]

Step 3: Calculate the partial pressures of each gas.

• For CH₄

[tex]\begin{aligned} P_{\text{CH}_4} & = X_{\text{CH}_4}P_{\text{T}} \\ & = \text{(0.53963)(28.3 atm)} \\ & = \boxed{\text{15.3 atm}} \end{aligned}[/tex]

• For Ar

[tex]\begin{aligned} P_{\text{Ar}} & = X_{\text{Ar}}P_{\text{T}} \\ & = \text{(0.11851)(28.3 atm)} \\ & = \boxed{\text{3.35 atm}} \end{aligned}[/tex]

• For Ne

[tex]\begin{aligned} P_{\text{Ne}} & = X_{\text{Ne}}P_{\text{T}} \\ & = \text{(0.34186)(28.3 atm)} \\ & = \boxed{\text{9.67 atm}} \end{aligned}[/tex]

Hence, the partial pressures of CH₄, Ar, and Ne are 15.3 atm, 3.35 atm, and 9.67 atm, respectively.

[tex]\\[/tex]

#CarryOnLearning

18. 1. A balloon has a pressure of 2.05 atm and at 28°C. If the temperature is reduced to 16°C, what will be the new pressure of the balloon? a. 3.97 atm b. 1.97 atm c. 2.97 atm d. 4.97 atm

Answer:

b. 1.97 atm

Explanation:

[tex]\frac{2.05(273+16)}{273+28} = 1.96827 atm = 1.97 atm[/tex]

19. A gas has an initial pressure and temperature of 70 atm and 350 K, respectively. What is the pressure of a gas at 500 K? a. 200 atm b. 49 atm c. 88 atm d. 100 atm

solution:
[tex] \frac{p1}{p2} = \frac{t1}{t2} \\ \frac{71 \: atm}{p2} = \frac{350k}{500k} \\ p2 = \frac{71atm}{ \frac{350k}{500k} } \\ p2 = 101.4 \: atm[/tex]
so the answet is letter D. which is the closest.

Pls give me feedback! thanks

20. each of the following containers has same size. which of following containers has the most compressed gas molecules? A. 10 atm B. 8 atm C. 6 atm D 4 atm

If all the containers have the same size. the answer must be A.A. 10 ATMOSPHERES becoz you will know it

21. Affidavit of loss atm card landbank was not return to atm machine

answer:

way kind of question is that

Explanation:

sorry po(I can help you na mn)but pls ipa gets nio po sa comment section

Answer:

sorry po sorry need na need kolang po kung may magdadagot po follow ko :(

22. what are the advantage of having atm card with you always?​

Answer:

Eliminate Cash – You don't have to carry around cash if you regularly use a debit card. If you do need cash you can always stop by an ATM and take out however much you need or get cash back when you make a purchase at most stores.

Explanation:

Hopes it's help:)

23. A container holds 600 mL of Helium gas at a constant temperature and a pressure of 1.8 atm. What will be the pressure if the volume decreased to 350 mL? A. 3.58 atm B. 3.09 atm C. 0.8 atm D. 1.96 atm

SOLUTION:

Step 1: List the given values.

[tex]\begin{aligned} P_1 & = \text{1.8 atm} \\ V_1 & = \text{600 mL} \\ V_2 & = \text{350 mL} \end{aligned}[/tex]

Step 2: Calculate the final pressure by using Boyle's law.

[tex]\begin{aligned} P_2 & = \frac{P_1V_1}{V_2} \\ & = \frac{(\text{1.8 atm})(\text{600 mL})}{\text{350 mL}} \\ & = \boxed{\text{3.09 atm}} \end{aligned}[/tex]

Hence, the answer is B. 3.09 atm

[tex]\\[/tex]

#CarryOnLearning

24. what does atm mean not the atm in bank​

Answer:

ATM-Asynchronous Transfer Mode

25. The pressure of a gas is 750 torr when its volume is 400 mL. Calculate the pressure (atm) if the gas is allowed to expand to 600 mL at constant temperature. a. 0.66 atm b. 1.48 atm c. 500 atm d. 1125 atm

Answer:

The answer is letter A.

Explanation:

Given:

P1 = 750 torr

V1 = 400 mL

V2 = 600 mL

P2 =?

Convertion:

1 atm

750 torr x ________ = 0.99 atm (rounded off)

760 torr

1 L

400 mL x ________ = 0.4 L

1000 mL

1 L

600 mL x ________ = 0.6 L

1000 mL

Formula:

V1 P1 = V2 P2

V1 P1

P2 = _____

V2

Solution:

(0.4 L)(0.99 atm)

P2 = ________________

0.6 L

0.39 L atm

P2 = __________

0.6 L

P2 = 0.66 atm

26. 7. How many moles of nitrogen are present in the problem?A. 1.05B. 1.06C. 1.07D. 1.088.What is the total moles of the mixture?A. 1.60B. 1.61C. 1. 62D. 1.639. What is the total pressure of the mixture in terms of atm?A. 3.60B. 3.61C. 3.62D. 3.6310. By rearranging the formula for Dalton's law, what is the partial pressure ofoxygen?A. 1.20 atmB. 1.21 atmC. 1.22 atmD. 1.23 atm11. What is the calculated partial pressure of nitrogen?A. 2.40 atmB. 2.41 atmC. 2.42 atmD. 2.43 atmFor items 12-15, refer to this problem: In a laboratory experiment, 98 gof oxygen and 140 g nitrogen are combined in a 5.0 L at 298 K.Calculate for partial pressure of each gas and its total pressure.12. How many moles of oxygen are in the mixture?A. 1.5 molB. 2.5 molC. 3.5 molD. 4.5 mol13. In the problem above, what is the partial pressure of oxygen gas?A. 17.09 atm B. 17.10 atmC. 17.11 atmD. 17.12 atm14. According to the problem, what is partial pressure exerted by a 140 gnitrogen?A. 24.45 atm B. 24.50 atm C. 24.55 atm D. 24.60 atm15. From the given situation, what will be the total pressure of the mixture?A. 41.47 atm B. 41.57 atm C. 41.67 atm D. 41.77 atm​

Answer:

7.a

8.b

9.d

10.d

11.b

12.a

13.d

14.a

15.c

27. 5. What is the Standard Temperature and Pressure (STP)? A. 1 K, 273 atm B. 100 K, 2 atm C. 273 K 1 atm D. 373 K 1 atm​

Answer:

1 atm at 273 k

Explanation:

Answer:

mali yung nasa taas

Explanation:

eto tama sagot wala ako pake

28. Which of the following refers to standard temperature and pressure? 0 atm and 273 K1 atm and 273 K0 atm and 273 degrees C1 atm and 273 degrees C​

Answer:

1 atm at 273 k

Explanation:

yan Ang pagkakaalam ko

29. At 46C a sample of ammonia gas exerts a pressure of 5.3 atm. What is the pressure when the volume of the gas is reduced to one-tenth (0.10) of the original value at the same temperature? a.53 atm b.71 atm c.45 atm d. 67 atm

Answer:

The answer is A. 53 atm

Explanation:

#CarryonLearningAnswer:

A. 53 atm

Solution:46 degrees=273+47k=320k=t1=t2Let volume1 of gas be vPressure1 = 5.3atmFor second situationTemperature is sameVolume2=1/10vPressure2=?P1v1/t1=p2v2/t2=5.3*v=?*1/10v= 53 atm

Hope that this will help you

#Trust Me

30. 4. A sample of neon has a volume of 225 cm at 2 atm of pressure. What will be the pressure of the gas to have a volume of 500 cm? a. 0.2 atm c. 1 atm b. 0.9 atm d. 4 atm​

Answer:

Hence, the pressure of the gas that have a volume of 500cm will be 0.9atm (B.)

Step-by-step explanation:

To find out what will be the pressure of a gas that has a volume of 500cm, we must use the Boyle's Equation:

[tex] \boxed{\bold{P_1V_1 = P_2V_2}}[/tex]

Since we're finding to the new pressure, we can rewrite the equation:

[tex] \boxed{\bold{P_2 = \frac{V_1P_1}{V_2}}}[/tex]

Substitute the given to the formula

[tex] P_2 = \frac{(225)(2)}{500} [/tex][tex] P_2 = \frac{450}{500} [/tex][tex] P_2 \approx 0.9 [/tex]

Hence, the pressure of the gas that have a volume of 500cm will be 0.9atm (B.)