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3 years ago

How much would a 50-kg object weigh on Mercury (gravity on Mercury is 3.59 m/s2)?

1 answer:

4 0

Gravitational Acceleration or W=Mg so... Q1: W(earth)=50, W(X)=500 ---> g(X)=10*g(earth)=10*9.8=98 --> C is correct. Q2: M=W/g --> M=735/9.8=75 ---> B is correct. Q3: W=Mg=50*3.59=179.5 N ---> B is correct

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Describe in your own words, in terms of particle movement and energy, what occurs when a liquid is heated to its boiling point.

tino4ka555 [31]

The atoms or molecules attain enough kinetic energy to overcome any intermolecular attractions they have. Since there are no longer any attractive forces between the particles, they are free to drift away into space. The same sort of thing happens in ordinary evaporation, but only at the surface.

5 0

3 years ago

According to the Arrhenius concept, which of the following substances is NOT a base in aqueous solutionsA. NH3B. LiOHC. NaOHD. H

olga_2 [115]

Answer:

NH₃ (Option A)

Explanation:

Arrhenius theory explained that the acids are the ones that have H⁺, either H in its formula. Following this, the bases are the ones that have OH⁻ , either OH and its formula.

It can be used only with compounds with H, or OH.

So the ammonia is not a base, as Arrhenius theory.

It is known that ammonia behaves as a weak base, but it does not have hydroxide ions that can yield to water

5 0

4 years ago

A sample of neon has a volume of 40.81 m3 at 23.5C. At what temperature, in Kelvins, would the gas occupy 50.00 cubic meters? As

mezya [45]

At $\fbox{\begin \\363 K \end{minispace}}$ temperature, a sample of neon gas will occupy $50.00 \text{ m}^{3}$ volume.

Further Explanation:

The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”

Mathematically the law can be expressed as,

$\fbox{ \begin \\ V \propto T \end{minispace}}$

Or,

$\frac{V}{T}=k$

Here, V is the volume of the gas, T is Kelvin temperature, and k is proportionality constant.

Given information:

The initial volume of neon gas is $40.81 \text{ m}^{3}$ .

The final volume of neon gas is $50.00 \text{ m}^{3}$ .

The initial temperature value is $23.5 \text{ } ^{\circ} \text{C}$ .

To calculate:

The final temperature

Given Condition:

The pressure is constant.

Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

Here,

$V_{1}$ is the initial volume of the gas.

$V_{2}$ is the final volume of the gas.

$T_{1}$ is the initial temperature of the gas.

$T_{2}$ is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

$\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}$ …… (2)

Step 3: Convert the given temperature from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

$T(\text{K}) = T(^{\circ}\text{C}) + 273.15$ …… (3)

Substitute $23.5\text{ }^{\circ} \text{C}$ for $T(^{\circ}\text{C})$ in equation (3) to convert temperature from degree Celsius to Kelvin.

$T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}$

Step 4: Substitute $40.81 \text{ m}^{3}$ for $V_{1}$ , $50.00 \text{ m}^{3}$ for $V_{2}$ and $296.65 \text{ K}$ for $T_{1}$ in equation (2) and calculate the value of $T_{2}$ .

$T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}$

Important note:

The temperature must be in Kelvin.

The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

5 0

3 years ago

Read 2 more answers

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.

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

A solution labeled "0.105 M NaOH" would contain ______________ moles of NaOH in each liter of solution.

koban [17]

It would contain 0.105M moles of NaOH in each liter of solution

5 0

3 years ago