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3 years ago

I WILL GIVE 15 POINTS PLS I NEED THIS BEFORE THE CLASS FINISHES IN 30 MINN

Chemistry

1 answer:

bulgar [2K]3 years ago

7 0

Answer:

likely be the same

Explanation:

this is because we have one color that both atoms share (green). both sample 1 and sample 2 have green and another color. yet, since they share one color, they are likely similar

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2NOCl(g) ↔ 2NO(g) + Cl2(g) with K = 1.6 x10–5. In an experiment, 1.00 mole of pure NOCl and 1.00 mole of pure Cl2 are placed in

poizon [28]

Answer: $3.8\times 10^{-3}M$

Explanation:

Moles of $NOCl$ = 1 mole

Moles of $Cl_2$ = 1 mole

Volume of solution = 1 L

Initial concentration of $NOCl$ = 1 M

Initial concentration of $Cl_2$ = 1 M

The given balanced equilibrium reaction is,

$2NOCl(g)\rightleftharpoons 2NO(g)+Cl_2(g)$

Initial conc. 1 M 0M 1 M

At eqm. conc. (1-2x) M (2x) M (1+x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[NO]^2[Cl_2]}{[NOCl]^2}$

The $K_c$ = $1.6\times 10^{-5}}$

Now put all the given values in this expression, we get :

${1.6\times 10^{-5}}=\frac{(2x)^2\times (1+x)}{(1-2x)^2}$

By solving the term 'x', we get :

$x=0.0019$

Concentration of $NO$ at equilibrium= (2x) M = $2\times 0.0019=3.8\times 10^{-3}M$

6 0

3 years ago

Assume that the density and heat of combustion of E85 can be obtained by using 85 % of the values for ethanol and 15 % of the va

Alika [10]

Answer:

87.4 J

Explanation:

The density of the gasoline is 0.70 g/mL, and the density of the ethanol is 0.79 g/mL. The heat combustions (the heat released in a combustion reaction) are 5,400 kJ/mol for gasoline, and 1,370 kJ/mol for ethanol.

For 3.5 L of E85, the volumes of gasoline and ethanol are:

Vgasoline = 0.15 * 3.5 = 0.525 L = 5.25x10⁻⁴ mL

Vethanol= 0.85 * 3.5 = 2.975 L = 2.975x10⁻³ mL

The mass of gasoline and ethanol presented in that sample of E85 is the volume multiplied by the density:

mgasoline = 5.25x10⁻⁴ * 0.70 = 3.675x10⁻⁴ g

methanol = 2.975x10⁻³ * 0.79 = 2.35025x10⁻³ g

The number of moles for each substance is it mass divided by its molar mass. The molar masses are 114 g/mol for gasoline, and 46 g/mol for ethanol:

ngasoline = 3.675x10⁻⁴/114 = 3.224x10⁻⁶ mol

nethanol = 2.35025x10⁻³ /46 = 5.109x10⁻⁵ mol

The energy released is the heat combustion multiplied by the number of moles, so:

Egasoline = 5,400 * 3.224x10⁻⁶ = 0.0174 kJ = 17.4 J

Eethanol = 1,370 * 5.109x10⁻⁵ = 0.07 kJ = 70 J

So, the energy released by the E85 is the sum of the energy released by ethanol and gasoline:

The energy released by E85 = 87.4 J

8 0

3 years ago

A sample of sandstone consists of only silica (SiO2) and calcite (CaCO3). When sandstone is heated, calcium carbonate decomposes

anzhelika [568]

Answer:

Explanation:

To calculate, the mass of Carbondioxide is 3.45mg

Then the decomposition of Carbondioxide gives 0 on calculation

5 0

4 years ago

Evan was in his bedroom when he heard the sound of breaking glass. He went out to the kitchen to see a vase broken on the floor

emmainna [20.7K]

The answer is d because he did not see the dog do it

7 0

3 years ago

Read 2 more answers

An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f

Dima020 [189]

The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

Mass of carbon, $C$ :

$\frac{32.18}{100}\times 223.94 = 72.06 g$

Mass of hydrogen, $H$ :

$\frac{4.5}{100}\times 223.94 = 10.08 g$

Mass of chlorine, $Cl$ :

$\frac{63.32}{100}\times 223.94 = 141.79 g$

Now, the number of each element in the unknown compound is determined by the formula:

$number of moles = \frac{given mass}{molar mass}$

Number of moles of $C$ :

$number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole$

Number of moles of $H$ :

$number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole$

Number of moles of $Cl$

$number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole$

Dividing each mole with the smallest number of mole, to determine the empirical formula:

$C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}$

$C_{1.5}H_{2.5}Cl_{1}$

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is $C_{3}H_{5}Cl_{2}$ .

Empirical mass = $12\times 3+1\times 5+2\times 35.5 = 112 g/mol$

In order to determine the molecular formula:

n = $\frac{molar mass}{empirical mass}$

n = $\frac{223.94}{112} = 1.99 \simeq 2$

So, the molecular formula is:

$2\times C_{3}H_{5}Cl_{2} = C_{6}H_{10}Cl_{4}$

6 0

3 years ago