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3 years ago

Water’s unique properties, high heat capacity, high density, solid phase less dense than liquid phase can be attributed to

Chemistry

1 answer:

Xelga [282]3 years ago

4 0

Answer:

B- The polarity of the molecules and hydrogen bonding between molecules.

Explanation:

Hope this helps:)

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Hydrogen cyanide, HCN, can be made by a two-step process. First, ammonia reacts with O2 to give nitric oxide, NO.

stealth61 [152]

Answer:

The mass of HCN is 79.65 g.

The mass of reactant which remain at the end of both reactions is 88.5 g.

Explanation:

Given that,

Mass of ammonia = 50.2 g

Mass of methane = 48.4 g

Hydrogen cyanide, HCN, can be made by a two-step process

Ammonia reacts with O₂ to give nitric oxide NO.

The reaction is,

$4NH_{3}+5O_{2}\Rightarrow 4NO+6H_{2}O$

We need to calculate the mole of NO

Using given data,

2.25 g NH_{3}=\dfrac{50.2}{17}= 2.95\ mole\ NH_{3} [/tex]

$4\ mole NH_{3}\ glose 4\ mol NO$

2.95 mol NH₃ will produced 2.95 mol NO

Then nitric oxide reacts with methane,

The reaction is,

$2NO+2CH_{4}\Rightarrow 2HCN+2H_{2}O+H_{2}$

We need to calculate the mole of methane

Using given data,

$mole\ of\ methane=\dfrac{48.4}{16}$

$mole\ of\ methane = 3.03\ moles$

2 mole NO produced 2 mole HCN

2.95 mol NO will produced $\dfrac{2.95\times3.03}{3.03}$ = 2.95 mol HCN

We need to calculate the mass of HCN

Using formula of mass

$m=N\times M$

Where, N = number of mole

M = molecular mass

Put the value into the formula

$m=2.95\times27$

$m= 79.65\ g$

The mass of HCN is 79.65 g.

We need to calculate the mass of NO

Using formula of mass

$m=N\times M$

Where, N = number of mole

M = molecular mass

Put the value into the formula

$m=2.95\times30$

$m= 88.5\ g$

Hence, The mass of HCN is 79.65 g.

The mass of reactant which remain at the end of both reactions is 88.5 g.

4 0

3 years ago

How many moles are in 24.75 g of H2O? What is given the conversation and the unknown?

riadik2000 [5.3K]

Answer:

1.373 mol H₂O

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

Explanation:

Step 1: Define

24.75 g H₂O

Step 2: Identify Conversions

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Step 3: Convert

$24.75 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O} )$ = 1.37347 mol H₂O

Step 4: Check

We are given 4 sig figs. Follow sig fig rules and round.

1.37347 mol H₂O ≈ 1.373 mol H₂O

7 0

2 years ago

How many moles would there be in 500,000,000,000 atoms of silver? Show your calculation

muminat

Answer: 3.01 * 10^35

Explanation:

500,000,000,000 * 6.02 * 10^23

4 0

3 years ago

Glucose, c6h12o6, is used as an energy source by the human body. the overall reaction in the body is described by the equation

Art [367]

Glucose is carbohydrate and a simple sugar that is very important to the human body.

Energy is produced for the cells in the body through the process of metabolism which oxidizes glucose to water, carbon dioxide, and some nitrogen compounds.

The general chemical reaction equation for metabolism is:

C6H12O6 + 6O2 ---> 6CO2 + 6H2O

4 0

3 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.

Back to this question. $[\text{H}^{+}]$ is given. 25 °C implies that $K_w = 10^{-14}$ . As a result,

$\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago