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alexgriva [62]
3 years ago
11

A weather balloon contains 1.10X10 to the power of 5 mol of helium and has a volume of 2.70x10 to the power of 6 L at 1.00 atm p

ressure. Calculate the temperature of the helium
Chemistry
1 answer:
olchik [2.2K]3 years ago
4 0

Answer:

299.14 K or 26°C

Explanation:

The ideal gas law, also called the general gas equation, is the equation of state of a hypothetical ideal gas.

The ideal gas law is often written as

PV = nRT

where P ,V and T are the pressure, volume and absolute temperature;

n is the number of moles of gas and  R is the ideal gas constant.

n=1.10 x 10^5 mol  

V= 2.70 x 10^6 L  

P= 1.00 atm= 101.325 kPa  

R= 8.314 kPa*L/ mol*K

when the formula is  rearranged, T=PV/ nR  

T = (101.325kPa * 2.70 x 10^6 L)/ (1.10 x 10^5 mol * 8.314 kPa*L/ mol*K)  

T = 299.1421917 K

or

T = 299.14 - 273.15 = 25.99 =  26°C

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F = 13 N

<h3>Further explanation</h3>

Given

Acceleration= a = 2.6 m/s²

Required

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Solution

Newton's second law :

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The wheeled cart used in the experiment is the same, so the mass used is the same

We use one of the data for example data 1 to find mass :

F = 1 N, a = 0.2 m/s²

m = F/a

m = 1/0.2

m = 5 kg

so for acceleration= a = 2.6 m/s² :

F = 5 kg x 2.6 m/s²

F = 13 N

8 0
2 years ago
Hat is the definition for a scientific theory? A. an informed explanation that can be tested but is not widely accepted B. an ex
frosja888 [35]

Answer:

D. a scientific explanation that is agreed on by all scientists and is no longer adjusted as new evidence is found

Explanation:

Scientific theory is an explanation that has been repeatedly tested and verified in accordance with the scientific method, using accepted protocols of observation, measurement, and evaluation of results

3 0
3 years ago
Read 2 more answers
1) The heat of combustion for the gases hydrogen, methane and ethane are −285.8, −890.4 and −1559.9 kJ/mol respectively at 298K.
Morgarella [4.7K]

Answer:

The enthalpy of the reaction is 64.9 kJ/mol.

Explanation:

H_2 + \frac{1}{2}O_2\rightarrow H_2O,\Delta H_1 =-285.8 kJ..[1]

CH_4 + 2O_2\rightarrow CO_2 + 2H_2O,\Delta H_2 =-890.4 kJ..[2]

C_2H_6 + \frac{7}{2}O_2\rightarrow 2CO_2 + 3H_2O,\Delta H_3= -1559.9 kJ..[3]

2CH_4(g)\rightarrow C_2H_6(g) + H_2(g),\Delta H_4=?..[4]

2 × [2] - [1]- [3] = [4]  (Using Hess's law)

\Delta H_4=2\times \Delta H_2 -\Delta H_1 -\Delta H_3

\Delta H_4=2\times (-890.4 kJ)-(-285.8 kJ) -(-1559.9 kJ)

\Delta H_4=64.9 kJ/mol

The enthalpy of the reaction is 64.9 kJ/mol.

3 0
3 years ago
Express the answer in scientific notation and with the correct number of significant figures: (6.32 x 10-4) ÷ 12.64
elixir [45]
<span>Express the answer in scientific notation and with the correct number of significant figures:
(6.32 x 10-4) ÷ 12.64
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3 years ago
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Sodium acetate can be formed from the metathesis/double replacement reaction of sodium
telo118 [61]

Answer:

Explanation:

Sodium Acetate Trihydrate BP Specifications

Sodium Acetate BP

C2H3NaO2,3H2O

Action and use

Used in solutions for dialysis; excipient.

DEFINITION

Sodium ethanoate trihydrate.

Content

99.0 per cent to 101.0 per cent (dried substance).

CHARACTERS

Appearance

Colourless crystals.

Solubility

Very soluble in water, soluble in ethanol (96 per cent).

IDENTIFICATION

A. 1 ml of solution S (see Tests) gives reaction (b) of acetates.

B. 1 ml of solution S gives reaction (a) of sodium.

C. Loss on drying (As shown in the Relevant Test).

TESTS

Solution S

Dissolve 10.0 g in carbon dioxide-free water prepared from distilled water R and dilute to 100 ml 100 ml with the same solvent.

Appearance of solution

Solution S is clear and colourless.

pH

7.5 to 9.0.

Dilute 5 ml of solution S to 10 ml with carbon dioxide-free water.

Reducing substances

Dissolve 5.0 g in 50 ml of water, then add 5 ml of dilute sulphuric acid and 0.5 ml of 0.002 M potassium permanganate. The pink colour persists for at least 1 h. Prepare a blank in the same manner but without the substance to be examined.

Chlorides

Maximum 200 ppm.

Sulphates

Maximum 200 ppm.

Aluminium

Maximum 0.2 ppm, if intended for use in the manufacture of dialysis solutions.

Arsenic

Maximum 2 ppm, determined on 0.5 g.

Calcium and magnesium

Maximum 50 ppm, calculated as Ca.

Heavy metals

Maximum 10 ppm.

Iron

Maximum 10 ppm, determined on 10 ml of solution S.

Loss on drying

39.0 per cent to 40.5 per cent, determined on 1.000 g by drying in an oven at 130C.

Sodium Acetate FCC Food Grade, US Food Chemical Codex

C2H3NaO2 Formula wt, anhydrous 82.03

C2H3NaO2·3H2O Formula wt, trihydrate 136.08

DESCRIPTION

Sodium Acetate occurs as colorless, transparent crystals or as a granular, crystalline or white powder. The anhydrous form is hygroscopic; the trihydrate effloresces in warm, dry air. One gram of the anhydrous form dissolves in about 2 mL of water; 1 g of the trihydrate dissolves in about 0.8 mL of water and in about 19 mL of alcohol.

Function: Buffer.

REQUIREMENTS

Identification: A 1:20 aqueous solution gives positive tests for Sodium and for Acetate.

Assay: Not less than 99.0% and not more than 101.0% of C2H3NaO2 after drying.

Alkalinity Anhydrous: Not more than 0.2%; Trihydrate: Not more than 0.05%.

Lead: Not more than 2 mg/kg.

Loss on Drying: Anhydrous: Not more than 1.0%; Trihydrate: Between 36.0% and 41.0%.

Potassium Compounds: Passes test.

5 0
3 years ago
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