A sample of water has a volume of 24.0 millimeters and a mass of 23.8 grams. what is the density

Please help its easy but i don't feel like doing it

KonstantinChe [14]

Answer:

first one is a second one is e

Explanation:

7 0

3 years ago

Predict how many H1 NMR signals (individual resonances, not counting splitting) are expected for the compound.

Lyrx [107]

Answer:

3 H1 NMR signals

Explanation:

NB: kindly check the diagram of the chemical compound in the attached picture.

This particular Question is based on the part of chemistry which is known as spectroscopy. Spectroscopy is used in the Determination or in identifying chemical compounds. H'NMR works on the principle of nuclear magnetic resonance.

In order to solve this question, one has to count the number of hydrogen in unique location. The diagram in the attached show how hydrogen is been counted.

The numbers of signals is the number of different chemical environments in which hydrogen atoms are located.

NB: signals is also the same as peak in H'NMR.

Hence, the number of H1 NMR signals in this chemical compound is 3.

3 0

3 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

Plz help me guys A B C D

Taya2010 [7]

Answer:

C

Explanation:

I think

3 0

4 years ago

C10H12O4S(s) + O2(g)  CO2(g) + SO2(g) + H2O(g)

ale4655 [162]

Answer: The coefficient for $O_2(g)$ is 12.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$C_{10}H_{12}O_4S(s)+12O_2(g)\rightarrow 10CO_2(g)+SO_2(g)+6H_2O$

Thus in the reactants, there are 12 molecules of oxygen in balanced chemical equation. Thus the coefficient for $O_2(g)$ is 12.

8 0

3 years ago