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Sophie [7]
3 years ago
7

Given the following sets of values, calculate the unknown

Chemistry
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

3.91 L

Explanation:

Using the ideal gas law equation as follows:

PV = nRT

Where:

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

Based on the information given in this question,

P = 5.23 atm

V= ?

n = 0.831 mol

T = 27°C = 27 + 273 = 300K

Using PV = nRT

V = nRT/P

V = (0.831 × 0.0821 × 300) ÷ 5.23

V = 20.47 ÷ 5.23

V = 3.91 L

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3. How would you expect the density of the gummy bear to change if you soaked it in isopropanol (rubbing alcohol, density
77julia77 [94]

Change occurs in the density of the gummy bear to change if we soaked it in isopropanol.

<h3>Effects on density by soaking isopropanol</h3>

We expect the density of the gummy bear to change if we soaked it in isopropanol because the gummy bear absorb the liquid which increase its density due to increasing weight of the gummy bear.

If a substance gains weight then its density also increases and we know that when the gummy bear was soaked in the isopropanol then its weight increases so its density is also changes so we can conclude that change occurs in the density of the gummy bear to change if we soaked it in isopropanol.

Learn more about density here: brainly.com/question/1354972

7 0
2 years ago
Calculate: (a) the weight (in lbf) of a 30.0 lbm object. (b) the mass in kg of an object that weighs 44N. (c) the weight in dyne
belka [17]

Answer:

a) 965,1 lbf

b) 4,5 kg

c) 1,33 * 10^6 dynes

Explanation:

Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.

Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.

                                        w=mg

In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International  System) or 32,17 ft/s² (in the FPS system).

To solve this problem we'll use the following conversion factors:

1 lbf = 1 lbm*ft/s²

1 N = 1 kg*m/s²

1 dyne = 1 gr*cm/s²   and 1 N =10^5 dynes

1 ton = 907,18 kg

1 k = 1000 gr

a) m = 30 lbm

w = 30 lbm * 32,17 ft/s^{2} = 965, 1 \frac{lmb*ft}{s^{2} } = 965,1 lbf

b) w = 44 N

First, we clear m of the weight equation and then we replace our data.

m = \frac{w}{g} = \frac{44 N}{9,8 \frac{m}{{s}^{2}} } = 4,5 kg

c) m = 15 ton

m=15 ton * \frac{907,18 kg}{1 ton} = 13607,7 kg \\ w = mg = 13607,7 kg * 9,8 m/s2 = 133355,5 N * \frac{10^{5} dynes }{1 N} = 1,33 * 10^{6}dynes

4 0
4 years ago
2. A gas occupies 320. ml at a pressure of 420.5 mmHg. Determine the volume if the pressure is decreased to 300. mmHg.
arsen [322]

Explanation:

v1 p1 = v2 p2

v2 = v1p1/p2

v2= 320×420.5/300

v2=449ml

5 0
3 years ago
You are playing a game “will it float?” In this game, you are given a large, square can of tuna. If you know the density of wate
saul85 [17]

Answer:

I think it's just density of the tuna

5 0
3 years ago
A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

5 0
3 years ago
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