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3 years ago

15

Analysis of an unknown compound reveals that the compound is 27.29% carbon and 72.71% oxygen. The molar mass of the compound is

132.03g/mol. What is the molecular formula of the compound?

1 answer:

Nana76 [90]3 years ago

6 0

Answer:

qdxbiuqn;xjh

Explanation:

hbxjibwjd jck`,2ebox

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Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

<u>Answer:</u> The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

<u>Initial:</u> 1.30 1.65

<u>At eqllm:</u> 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

8 0

3 years ago

Read 2 more answers

What is the mass percent of oxygen (0) in SO2?

vazorg [7]

Answer:

D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

Explanation:

Step 1: Detemine the mass of O in SO₂

There are 2 atoms of O in 1 molecule of SO₂. Then,

m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g

Step 2: Determine the mass of SO₂

m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g

Step 3: Detemine the mass percent of oxygen in SO₂

We will use the following expression.

m(O)/m(SO₂) × 100%

(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%

5 0

3 years ago

I don't need to know why i just want the awnser

Lubov Fominskaja [6]

Answer:

I remember doing this in 7th,

1. D

2. B or D, more leaning on B though

3. A

4 0

3 years ago

Which general equation shows a Single-displacement reaction?

Stella [2.4K]

Hello! The answer is D

A good note for these is when there are three elements, one being a singular element and another a compound and there’s a single switch, this could show a single-displacement

Have a good day gamer.

6 0

2 years ago

2. How many grams of water can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories?

sertanlavr [38]

Answer:

672 g

Explanation:

We can calculate the mass of water that can be warmed from 25.0°C to 37.0°C by the addition of 8,064 calories using the following expression.

$Q = c \times m \times \Delta T$

where,

c: specific heat of the water

m: mass

ΔT: change in the temperature

$m = \frac{Q}{c \times \Delta T } = \frac{8,064cal}{(1cal/g. \° C) \times (37.0 \° C - 25.0 \° C) } = 672 g$

The mass of water that can be warmed under these conditions is 672 grams.

5 0

3 years ago