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2 years ago

In chemical reactions, how does the valence configuration of bromine tend to change?

1 answer:

3 0

Answer: Near the Protons. The electron structure of bromine is illustrated above. In chemical reactions, how does the valence configuration of Bromine tend to change? ... It loses one electron.

Explanation:

btw i found that on google lol

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A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol

Kryger [21]

Answer:

0.44 moles

Explanation:

Given that :

A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.

The equilibrium constant $K_c= \dfrac{[CO][H_2]}{[H_2O]}$

The equilibrium constant $K_c= \dfrac{(0.17 )(0.17)}{0.74}$

The equilibrium constant $K_c= 0.03905$

Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.

The equation for the reaction is :

$H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \ \ \ \ \ \to0.17$

Total mole of water now = 0.74+0.17

Total mole of water now = 0.91 moles

Again:

$K_c= \dfrac{[CO][H_2]}{[H_2O]}$

$0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}$

0.03905(0.91 -x) = (0.17 +x)(x)

0.0355355 - 0.03905x = 0.17x + x²

0.0355355 +0.13095 x -x²

x² - 0.13095 x - 0.0355355 = 0

By using quadratic formula

x = 0.265 or x = -0.134

Going by the value with the positive integer; x = 0.265 moles

Total moles of CO in the flask when the system returns to equilibrium is :

= 0.17 + x

= 0.17 + 0.265

= 0.435 moles

=0.44 moles (to two significant figures)

3 0

3 years ago

How much ch2o is needed to prepare 445 ml of a 2.65 m solution of ch2o?

worty [1.4K]

Answer: 35.4 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

Molality = 2.65

n= moles of solute =?

$V_s$ = volume of solution in ml = 445 ml

Putting in the values we get:

$2.65=\frac{n\times 1000}{445ml}$

$n=1.18$

Mass of solute in g= $moles\times {\text {molar mass}}=1.18mol\times 30.02g/mol=35.4g$

Thus 35.4 grams of $CH_2O$ is needed to prepare 445 ml of a 2.65 m solution of $CH_2O$ .

8 0

2 years ago

Meanders are characteristics of a young river. True False

Rainbow [258]

True would be the answer

7 0

3 years ago

What functions of cells can be affected by a mutation?

Colt1911 [192]

It changes the rate of growth that cells usually undergo.

6 0

2 years ago

Convert 6.7 x 1024 molecules of nitrogen dioxide into grams.

BlackZzzverrR [31]

Answer:

510 g NO₂

General Formulas and Concepts:

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Reading the Periodic Table
Writing Compounds
Using Dimensional Analysis

Explanation:

Step 1: Define

6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)

Step 2: Define conversions

Avogadro's Number

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol

Step 3: Use Dimensional Analysis

$6.7 \cdot 10^{24} \ molecules \ NO_2(\frac{1 \ mol \ NO_2}{6.022 \cdot 10^{23} \ molecules \ NO_2} )(\frac{46.01 \ g \ NO_2}{1 \ mol \ NO_2} )$ = 511.901 g NO₂

Step 4: Check

We are given 2 sig figs. Follow sig fig rules.

511.901 g NO₂ ≈ 510 g NO₂

6 0

2 years ago