convert 20.8 atm to kPa
1. convert 20.8 atm to kPa
Answer; 23.0
Step-by-step explanation:
2. Which of the following is the correct order of pressures in increasing magnitude? A. 410 mmHg, 0.60 atm, 49 torr, 70 kPa B. 49 torr, 410 mmHg, 0.60 atm, 70 kPa C. 70 kPa, 0.60 atm, 410 mmHg, 49 torr D. 0.60 atm, 49 torr, 70 kPa, 410 mmHg
Answer:
B. 49 torr, 410 mmHg, 0.60 atm, 70 kPa
3. convert 1.8 atm to Pa and kPa
Answer:
1.8 atm = 182385 pa
1.8 atm = 182.385 kpa
Explanation:
4. which of the following measurements shows the smallest air pressure? A. 1.00 atm B. 76mmHg C. 101 Kpa D. 760cmHg
Answer:D. 760 mmHg
:)
Answer:
D
Explanation:
5. 0.955 atm to kPa. help me plss
Answer:
wala sad ko ka baw check your module
6. What is the pressure in atm if the pressure is equal to 7500 KPa?
Answer:
74.019 atm
Explanation:
1 kPa is equal to 0.009869 atm
7. which of the following is equal to the standard pressure?a. 1 atmb. 1 torrc. 273 kPad. 760 atm(ang nonsense, report.)
Answer:
a. 1 atm
[tex] \\ \\ [/tex]
Answer:
A. 1 atm
Explanation:
Standard pressure is 1 atm. It can also be 101.3kPa or 760 mmHg.
#CarryOnLearning
8. A sample of neon gas occupies 0.220 L at 0.860 atm. What will be its volume at 29.2kpa pressure? 1 atm = 101.3 kpa
initial volume=0.220 L
initial pressure =0.860 atm
final volume = ?
final pressure = 29.2 kpa(0.288 atm)
v2=P1v1/P2
=0.860 atm (0.220 L) / 0.288 atm
=0.657 L
P=⬇
V=⬆
P inversly proportional 1/v
9. How many liters will 0.20 mol of hydrogen iodide at 300 K and 100.0 kPa occupy? Use R = 8.314 kPa*L/(K*mol) = 0.08205 atm*L/(mol*K).
Answer:
Volume = 4.988 L
Explanation:
PV = nRT
V = nRT
P
V = (0.2 mol) (8.314 kPa*L/K*mol) (300 K)
100 kPa
V = 4.988 L
or
convert kPa to atm
V = (0.2 mol) (0.08205 atm*L/K*mol) (300 K)
0.9869 atm
V = 4.988 L
10. which is referred to as the standard pressure?a. 1 kPa b. 1 psic. 760 atmd. 760 torr
Answer:
============
A.
============
Explanation:
#keepONlearning
hmmm tama ba yung # ko?
11. A sample of a gas has a volume of 20L at 1.5 atm. What will be the volume of this gas at eachof the following pressure:a.700 mmHgb. 2 atmc.105 kPa
a. 700 mmHgGiven:
[tex]P_{1} = \text{1.5 atm} × \frac{\text{760 mmHg}}{\text{1 atm}} = \text{1140 mmHg}[/tex]
[tex]V_{1} = \text{20 L}[/tex]
[tex]P_{2} = \text{700 mmHg}[/tex]
Required:[tex]V_{2}[/tex]
Equation:[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
Solution:[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
[tex]V_{2} = V_{1} × \frac{P_{1}}{P_{2}}[/tex]
[tex]V_{2} = \text{20 L} × \frac{\text{1140 mmHg}}{\text{700 mmHg}}[/tex]
[tex]\boxed{V_{2} = \text{33 L}}[/tex]
Answer:[tex]V_{2} = \text{33 L}[/tex]
b. 2 atmGiven:[tex]P_{1} = \text{1.5 atm}[/tex]
[tex]V_{1} = \text{20 L}[/tex]
[tex]P_{2} = \text{2 atm}[/tex]
Required:[tex]V_{2}[/tex]
Equation:[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
Solution:[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
[tex]V_{2} = V_{1} × \frac{P_{1}}{P_{2}}[/tex]
[tex]V_{2} = \text{20 L} × \frac{\text{1.5 atm}}{\text{2 atm}}[/tex]
[tex]\boxed{V_{2} = \text{15 L}}[/tex]
Answer:[tex]V_{2} = \text{15 L}[/tex]
c. 105 kPaGiven:[tex]P_{1} = \text{1.5 atm} × \frac{\text{101.325 kPa}}{\text{1 atm}} = \text{151.9875 kPa}[/tex]
[tex]V_{1} = \text{20 L}[/tex]
[tex]P_{2} = \text{105 kPa}[/tex]
Required:[tex]V_{2}[/tex]
Equation:[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
Solution:[tex]P_{1}V_{1} = P_{2}V_{2}[/tex]
[tex]V_{2} = V_{1} × \frac{P_{1}}{P_{2}}[/tex]
[tex]V_{2} = \text{20 L} × \frac{\text{151.9875 kPa}}{\text{105 kPa}}[/tex]
[tex]\boxed{V_{2} = \text{29 L}}[/tex]
Answer:[tex]V_{2} = \text{29 L}[/tex]
#CarryOnLearning
12. Answer the following1. Checking a barometer Gian measured the height of Mercury and found it to be 72.8 cm. Report this value in atm, torr, and pascals.2.On a certain day, the atmospheric pressure in Davao city was 759 mmHg what was the pressure in atm, and kPa (1 kilopascal=1000pascals)3. The pressure of an automobile tire was reported as 2.68 X 10^5 kPa. Determine the equivalent value in atm and psi.
dSo yeah I don't really know solve it go motivate yourself
13. ACTIVITY 1 Convert the following pressure and temperature: 1. 45 mmHg - atm 2. 75 atm - torr 3. 1456 kPa - atm 4. 456 K-°C 5. 67 °C-K
Answer:
10. Alin sa mga sumusunod ang isa sa naging mabisang paraan na nagpakita ng malawakangpagkakaisa ng mamamayan at makamit ang kalayaan mula sa rehimeng Marcos?A. Pagsaagawa ng hunger strikeB. Paggawa ng pelikulang na nakasentro sa isyung panlipunanC. Paghikayat sa mga simbahan upang magsagawa ng sabayangpananalanginD. Malawakang pagdalo sa EDSA upang makiisa sa protesta sa pamamagitan ng pagdarasal, pag-awit,pakikinig sa mga talumpati at pakikipag-usap sa mga sundalong nakatalagaExplanation:
10. Alin sa mga sumusunod ang isa sa naging mabisang paraan na nagpakita ng malawakangpagkakaisa ng mamamayan at makamit ang kalayaan mula sa rehimeng Marcos?A. Pagsaagawa ng hunger strikeB. Paggawa ng pelikulang na nakasentro sa isyung panlipunanC. Paghikayat sa mga simbahan upang magsagawa ng sabayangpananalanginD. Malawakang pagdalo sa EDSA upang makiisa sa protesta sa pamamagitan ng pagdarasal, pag-awit,pakikinig sa mga talumpati at pakikipag-usap sa mga sundalong nakatalagaAnswer:
1. 0.059 or 0.060 atm
2. 2090 torr
3. 14.3731 atm
4. 182.85 °C
5. 340.15 K
Explanation:
1. 45 mmHg to atm
45 mmHg · 1 atm ÷ 760 mmHg = 0.059 or 0.060 atm
2. 75 atm to torr
75 atm · 760 torr ÷ 1 atm = 2090 torr
3. 1456 kPa - atm
1456 kPa · 1 atm ÷ 101.3 kPa = 14.3731 atm
4. 456 K to Celcius
K - 273.15 = °C
456 K - 273.15 = 182.85 °C
5. 67 °C to Kelvin
°C + 273.15 K
67 + 273.15 K = 340.15 K
14. Which conditions are equal to STP? 30°C, 100 kPa –273.15°C, 1 atm 0°C, 760 mmHg 25°C, 101.325 kPa
Conditions are equal to STP : 0 ° C (273.15 K) and 1 atm (101,325 kPa = 760 mmHg)
Further explanationStandard conditions for temperature and pressure are used as a reference in certain calculations or conditions
There are several standards used by various organizations such as IUPAC or NIST
There are 2 conditions that are usually used as a reference in chemical calculations (mainly for determining the volume per mole of a gas or the molar volume), namely:
1. Standard ConditionsConditions at T 0 ° C (273.15 K) and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
2 Room ConditionConditions at T 25 ° C (298 K) and P 1 atm are stated by RTP (Room Temperature and Pressure). Vm in this condition = 24.4 liters / mol
So 1 atm if converted to Pascal/mmHg
1 atm = 101,325 kPa = 760 mmHg
So the STP condition is at 0 ° C (273.15 K) and 1 atm (101,325 kPa = 760 mmHg)
Learn more
the density of NH3 gas at STP
https://brainly.ph/question/2113522
the volume occupied by 0.25 mol of a gas at STP
https://brainly.ph/question/2139883
volume of gas occupy at STP
https://brainly.ph/question/2168937
https://brainly.ph/question/2131548
https://brainly.ph/question/1989131
https://brainly.ph/question/1321381
molecules in 30 liters of methane (CH4) at STP
https://brainly.ph/question/2141313
#LetsStudy
15. Convert 1.8 atm to Pa an kPaneed your help ASAP
Answer:
.1824Pa, 182.394Kpa
Explanation:
1atm = 101.33kpa
(1.8atm)x(101.33kpa/1atm)
=182.394kpa
182.394kpa(1kpa/1000Pa)
=.1824Pa
16. WORKSHEET #4: Conversion 1) 6.67 atm -> kPa 4) 639 mmHg -> tort 5) 8.7 atm --> Pa 2) 242 mmHg - atm 3) 54.3 kPa torr Pa help nmn po:>
such road ️ you
Explanation:
I can explain-
17. How much volume (in L) will 0.20 mol hydrogen iodide at 300 K and 100.0 kPa occupy? Use R = 8.314 kPa*L/(K*mol) = 0.08205 atm*L/(mol*K)
This question is under stoichiometry which is the quantitative relationship of reactants and products.
We will use the formula for ideal gas law which is PV=nRT
P = pressure, T = temperature, V = volume, n = # of moles, R = ideal gas constant
PV = nRT
in order to find the volume divide P on both sides of the equation, we get:
V = nRT / P (substitute the values given)
V = (0.20) (8.314) (300) / 100
*Make sure units are in kPA, mol, K if you want to use the 8.314
V = 4.9884 L
18. How much volume (in L) will 0.20 mol hydrogen iodide at 300 K and 100.0 kPa occupy? Use R = 8.314 kPa*L/(K*mol) = 0.08205 atm*L/(mol*K).
Use the formula for the ideal gas law which is PV=nRT
PV=nRT
To determine the volume divide P on both sides of the equation, we need to do the following:
V = nRT / P (Substitute the corresponding values)
V = (0.20) (8.314) (300) / 100
[ If you're going to use the 8.314, make sure the units are in kPA, mol, and K. ]
Answer: V = 4.9884 L
19. 1. The total pressure in a closed container of three mixed gases is 96.4 kPa. The partial pressure of hydrogen in the mixture is 13.5 kPa and the partial pressure of oxygen is 29.3 kPa. The third gas in the mixture is methane, what is its partial pressure? 2. Oxygen and chlorine gas are mixed in a container with partial pressures of 401 mmHg and 0.639 atm, respectively. What is the total pressure inside the container (in atm)?
Problem 1:Given:
[tex]P_{\text{T}} = \text{96.4 kPa}[/tex]
[tex]P_{\text{H}_{2}} = \text{13.5 kPa}[/tex]
[tex]P_{\text{O}_{2}} = \text{29.3 kPa}[/tex]
Required:[tex]P_{\text{CH}_{4}}[/tex]
Solution:[tex]P_{\text{T}} = P_{\text{H}_{2}} + P_{\text{O}_{2}} + P_{\text{CH}_{4}}[/tex]
[tex]P_{\text{CH}_{4}} = P_{\text{T}} - P_{\text{H}_{2}} - P_{\text{O}_{2}}[/tex]
[tex]P_{\text{CH}_{4}} = \text{96.4 kPa} - \text{13.5 kPa} - \text{29.3 kPa}[/tex]
[tex]\boxed{P_{\text{CH}_{4}} = \text{53.6 kPa}}[/tex]
[tex]\\[/tex]
Problem 2:Given:[tex]P_{\text{O}_{2}} = \text{401 mmHg} × \frac{\text{1 atm}}{\text{760 mmHg}} = \text{0.5276 atm}[/tex]
[tex]P_{\text{Cl}_{2}} = \text{0.639 atm}[/tex]
Required:[tex]P_{\text{T}} = P_{\text{O}_{2}} + P_{\text{Cl}_{2}}[/tex]
[tex]P_{\text{T}} = \text{0.5276 atm} + \text{0.639 atm}[/tex]
[tex]\boxed{P_{\text{T}} = \text{1.167 atm}}[/tex]
[tex]\\[/tex]
#CarryOnLearning
20. How to convert 1.8 atm to Pa and kPa?
Answer:
1.8 atm = 182385 Pa
1.8 atm = 182.385 kPa
Answer:
.1824pa,182.394 kpa
Explanation:
correcf me if I'm wrong
21. Convert 0.08205 atm to kPa, bar, and torr. Round of your answer to three significant figures.
Answer:
in order to avoid too much spilling of varnish,the brush should be dipped in the can at least a.¼of the length of the bristles b.1/4 of the length of the bristles c. ⅓ of the length of the bristles
22. What is the pressure in atm if the pressure is equal to 2300 KPa?
SOLUTION:
Note that 1 atm = 101.325 kPa.
[tex]\begin{aligned} ? \: \text{atm} & = \text{2300 kPa} \times \frac{\text{1 atm}}{\text{101.325 kPa}} \\ & = \boxed{\text{22.7 atm}} \end{aligned}[/tex]
Hence, the pressure in atm is 22.7 atm
[tex]\\[/tex]
#CarryOnLearning
23. At sample of gas volume of 20 L at 1.5 atm , what will be the volume of this gas each of the following pressure:a. 700 mmHgb. 2 atmc. 105 kPa
Answer:
C.YUNG ANSWER
Explanation:
Explanation:
This is what we're given:
V
1
, or initial volume, which is
2.5 L
.
P
1
, or initial pressure, which is
1.5 atm
.
P
2
, or final pressure, which is
1 atm
.
We have to find
V
2
, or the final volume.
Boyle's law actually relates all four of these variables!:
P
1
V
1
=
P
2
V
2
This equation can be rearranged to have only
V
2
on one side:
V
2
=
P
1
V
1
P
2
Now, we just need to plug in all of the variables to find
V
2
:
V
2
=
1.5 atm × 2.5 L
1 atm
V
2
=
1.5
atm
×
2.5 L
1
atm
V
2
=
3.75 L
PA BRAINLEST PLSSS
24. 1. WHAT HAPPENEDA. Test yourself: Converting gas pressure units.1. Convert 0.452 atm to torr.O2. Convert 7.9 x 10-2 torr to atm.3. Convert 247.2 to KPa to torr.4. Convert 0.456 atm to mm Hg.5. Convert 156.4 KPa to mm Hg.STEM
Solution (1)
[tex]P \: \text{(in torr) = 0.452 atm} × \frac{\text{760 torr}}{\text{1 atm}}[/tex]
[tex]\boxed{P \: \text{(in torr) = 343.5 torr}}[/tex]
Solution (2)
[tex]P \: \text{(in atm)} = 7.9 × 10^{-2} \: \text{torr} × \frac{\text{1 atm}}{\text{760 torr}}[/tex]
[tex]\boxed{P \: \text{(in atm)} = 1.04 × 10^{-4} \: \text{atm}}[/tex]
Solution (3)
[tex]P \: \text{(in torr) = 247.2 kPa} × \frac{\text{1 atm}}{\text{101.325 kPa}} × \frac{\text{760 torr}}{\text{1 atm}}[/tex]
[tex]\boxed{P \: \text{(in torr) = 1854 torr}}[/tex]
Solution (4)
[tex]P \: \text{(in mmHg) = 0.456 atm} × \frac{\text{760 mmHg}}{\text{1 atm}}[/tex]
[tex]\boxed{P \: \text{(in mmHg) = 346.6 mmHg}}[/tex]
Solution (5)
[tex]P \: \text{(in mmHg) = 156.4 kPa} × \frac{\text{1 atm}}{\text{101.325 kPa}} × \frac{\text{760 mmHg}}{\text{1 atm}}[/tex]
[tex]\boxed{P \: \text{(in mmHg) = 1173 mmHg}}[/tex]
#CarryOnLearning
25. convert a gas pressure of 191 ATM to kPa.
Answer:
Conversion factor: 1 Atm = 101.325 kPa1)
kPa = Atm * 101.3252)
kPa = 191 * 101.3253)
kPa = 19353.075
26. Express 1184 torr in units of mmHg, atm and kPa
1184 torr ( 760 mmHg / 760 torr ) =
➡️ 1184 mmHg
1184 torr ( 1 atm / 760 torr ) =
➡️ 1.557894737 atm
1184 torr ( 101.325 KPa / 760 torr ) =
➡️157.8536842 KPa
27. ConvenstenDirection Convert 1. 1 atm to torr2. 14.7 psi to kPa3. 760 mmHg to psi4. 725 torr to atm5. 35 psi to atm
1) 1 atm x (760 torr / 1 atm)
= 760 torr
2) 14.7 psi = 1 atm
1 atm = 101.325 kPa
3) 760 mmHg = 1 atm
1 atm = 14.7 psi
4) 725 torr x (1 atm / 760 torr) = 0.9539 atm
5) 35 psi x (1 atm / 14.7 psi) = 2.38 atm
28. (3 points) How many liters will 0.20 mol of hydrogen iodide at 300 K and 100.0 kPa occupy? Use R = 8.314 kPa*L/(K*mol) = 0.08205 atm*L/(mol*K
Answer:
4.9884 Liters of hydrogen iodide
Explanation:
Assuming ideal gas law applies, our working equation would be PV=nRT where P is the pressure, V will be the volume, n will be the number of moles, R is the gas constant (which is given in the problem), and T is the temperature.
The problem asks for the number of liters which means we need the volume. Rearranging the equation to get the volume, our final working equation will be V=[tex]\frac{nRT}{P}[/tex].
Substituting the given values V=[tex]\frac{(0.2)(8.314)(300)}{100}[/tex]
V= 4.9884 Liters
We used the value of R to be 8.314 kPa*L/K*mol for the reason of unit consistency.
More more information about ideal gas law you can visit this link https://brainly.ph/question/2081255
29. (3 points) How many liters will 0.20 mol of hydrogen iodide at 300 K and 100.0 kPa occupy? Use R = 8.314 kPa*L/(K*mol) = 0.08205 atm*L/(mol*K).
For this question, we can simply use the ideal gas law wherein PV=nRT. We can first transform the original equation into our working equation then substitute the values in the corresponding variables. Since all the parameters are already given except for volume which is what we are looking for in this problem we then have,
V = (nRT)/P
where, n=0.20 moles
T=300K
P=100kPa
R=8.314 kPa*L/K*mol
V = (0.20mol)(8.314kPa*L/K*mol)(300K)/100kPa.
All units will cancel except for L
= 4.9884
= 5.0 L
Therefore, 0.20 mol of hydrogen iodide at 300 K and 100.0 kPa will occupy 5.0L.
30. Express 1184 torr in units of mmHg, atm, and kPa
Answer:
di kopo gets ate/kuya
Explanation:
dikopo gets ate/kuya eh sorry po?
