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3 years ago

How does an element differ from a compound? How are they similar?

Chemistry

1 answer:

ArbitrLikvidat [17]3 years ago

7 0

Explanation:

DIFFERENCE:

The difference between element and compound is that the element is the substance which is made of same type of the atoms, whereas the compound is made of different elements which are combined in definite proportions.

SIMILARITY:

Elements and Compounds both contains atoms. They both have bonds linking atoms together. They both are pure substances and homogeneous in composition.

Examples of elements include copper, hydrogen, oxygen, etc.

Examples of compounds include water, salt NaCl, etc.

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A gas balloon has a volume of 80.0 mL at 300K, and a pressure of 50.0 kPa. If the pressure changes to 80 kPa and the temperature

fomenos

Answer: The new volume is 53.3 ml

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 50.0 kPa

$P_2$ = final pressure of gas = 80.0 kPa

$V_1$ = initial volume of gas = 80.0 ml

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $300K$

$T_2$ = final temperature of gas = $320K$

Now put all the given values in the above equation, we get:

$\frac{50.0\times 80.0}{300}=\frac{80.0\times V_2}{320}$

$V_2=53.3ml$

The new volume is 53.3 ml

4 0

2 years ago

At 25C the density of water is 0.997044 g/mL. Use this value to determine the percent error for the two density measurements

Gnom [1K]

Given that:

At 25C the density of water is 0.997044 g/mL.

From the information attached below, we have the following parameters.

The density of water calculation using a bottle.

Initial volume of Final volume of Mass of water Density (g/mL)

burette (mL) burette (mL) dispensed (g)

Sample 1 2.33 7.34 5.000 -----

Sample 2 7.34 12.37 5.025 -----

Sample 3 12.37 18.50 6.112 -----

Sample 4 18.50 24.57 6.064 -----

Sample 5 24.57 31.31 6.720 -----

The first thing we need to do is to determine the change in the volume of the burette in each sample from the above information.

The change in the volume of the burette = (final volume - the initial volume) mL

Sample 1:

= (7.34 - 2.33) mL

= 5.01 mL

Sample 2:

= (12.37 - 7.34) mL

= 5.03 mL

Sample 3:

= (18.50 - 12.37) mL

= 6.03 mL

Sample 4:

= (24.57 - 18.50) mL

= 6.07 mL

Sample 5:

= (31.31 - 24.57) mL

= 6.74 mL

The mass of the water dispersed in sample 1 is given as = 5.000 g

Using the relation for calculating the density of each, we have:

Sample 1

$\mathbf{density = \dfrac{mass}{volume}}$

$\mathbf{density = \dfrac{5.01 g}{5.000 ml}}$

density = 0.998004 g/ml

Sample 2:

$\mathbf{density = \dfrac{5.025 g}{5.03ml}}$

density = 0.999006 g/ml

Sample 3:

$\mathbf{density = \dfrac{6.112 g}{6.13ml}}$

density = 0.997064 g/ml

Sample 4:

$\mathbf{density = \dfrac{6.064 \ g}{6.07 \ ml}}$

density = 0.999012 g/ml

Sample 5:

$\mathbf{density = \dfrac{6.720 \ g}{6.74 \ ml}}$

density = 0.997033 g/ml

Thus, the average density for all the samples is:

$\mathbf{= \dfrac{( 0.998004 + 0.999006 + 0.997064 + 0.999012 + 0.997033 )}{5}}$

= 0.998024

∴

The percentage error for the two densities measurement is:

$=\dfrac{ (experimental \ value -theoretical \ value)\times 100 }{theoretical \ value}$

Given that the theoretical value = 0.997044 g/ml

Then;

$\mathbf{= \dfrac{(0.998024 - 0.997044)100}{0.997044}}$

= 0.0983%

Therefore, we can conclude that the percent error for the two density measurements is 0.0983%

Learn more about density here:

brainly.com/question/24386693?referrer=searchResults

4 0

2 years ago

A 1.24g sample of a hydrocarbon, when completely burned in an excess of O2 yields 4.04g Co2 and 1.24g H20. Draw plausible struct

arsen [322]

Answer:

Plausible structure has been given below

Explanation:

Molar mass of $CO_{2}$ is 44 g/mol and molar mass of $H_{2}O$ is 18 g/mol
Number of mole = (mass/molar mass)

4.04 g of $CO_{2}$ = $\frac{4.04}{44}moles$ $CO_{2}$ = 0.0918 moles of $CO_{2}$

1 mol of $CO_{2}$ contains 1 mol of C atom

So, 0.0918 moles of $CO_{2}$ contains 0.0918 moles of C atom

1.24 g of $H_{2}O$ = $\frac{1.24}{18}moles$ $H_{2}O$ = 0.0689 moles of $H_{2}O$

1 mol of $H_{2}O$ contain 2 moles of H atom

So, 0.0689 moles of $H_{2}O$ contain $(2\times 0.0689)moles$ of $H_{2}O$ or 0.138 moles of $H_{2}O$

Moles of C : moles of H = 0.0918 : 0.138 = 2 : 3

Empirical formula of hydrocarbon is $C_{2}H_{3}$

So, molecular formula of one of it's analog is $C_{4}H_{6}$

Plausible structure of $C_{4}H_{6}$ has been given below.

4 0

2 years ago

The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0

3 years ago

How many milliliters of 0.100 m naoh are required to neutralize 9.00 ml of 0.0500 m hcl?

BabaBlast [244]

V ( NaOH ) = mL ?

M ( NaOH ) = 0.100 M

V ( HCl ) = 9.00 mL / 1000 => 0.009 L

M ( HCl ) = 0.0500 M

number of moles HCl:

n = M x V

n = 0.009 x 0.0500 => 0.00045 moles HCl

mole ratio:

HCl + NaOH = NaCl + H2O

1 mole HCl ---------------- 1 mole NaOH
0.00045 moles HCl ----- ??

0.00045 x 1 / 1 => 0.00045 moles of NaOH

M = n / V

0.100 = 0.00045 / V

V = 0.00045 / 0.100

V = 0.0045 L

1 L ------------ 1000 mL
0.0045 L ----- ??

0.0045 x 1000 / 1 => 4.5 mL of NaOH

6 0

3 years ago