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3 years ago

What would be the mass of 3 moles of water molecules?

Chemistry

2 answers:

ankoles [38]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf A. \ 54 \ grams }}$

Explanation:

To convert from moles to grams, we must use the molar mass.

Recall that water's molecular formula is H₂O. It contains hydrogen and oxygen. Look up the two elements masses on the Periodic Table.

Hydrogen (H): 1.008 g/mol
Oxygen (O): 15.999 g/mol

Now, use these masses to find water's mass. The subscript of 2 tells us there are 2 atoms of hydrogen, so we multiply hydrogen's mass by 2 and add oxygen's.

H₂O= 2(1.008 g/mol) + 15.999 g/mol = 18.015 g/mol

Use the molar mass as a ratio.

$\frac{18.015 \ g \ H_2 O}{ 1 \ mol \ H_2 O}$

Multiply by the given number of moles.

$3 \ mol \ H_2O*\frac{18.015 \ g \ H_2 O}{ 1 \ mol \ H_2 O}$

The moles of water will cancel.

$3 *\frac{18.015 \ g \ H_2 O}{ 1 }$

$3 *{18.015 \ g \ H_2 O}$

$54.045 \ g \ H_2O$

Round to the nearest whole number. The 0 in the tenth place tells us to leave the number as is.

$54 \ g \ H_2O$

There are about <u>54 grams</u> of water in 3 moles.

frez [133]3 years ago

7 0

Answer:

54 grams

Explanation:

First step is to find what one mole of water is in grams.

Water is H2O so that means we have two moles of Hydrogen.

H - 2 grams (on periodic table)

We have one mole of Oxygen.

O - 16 grams (on periodic table)

Add 2 and 16 to get 18 total grams in one molecule of H2O

Now multiply 3 times 18 to know how many grams in 3 moles of H2O

You end with

3x18=54 grams

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A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

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In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

Regards!

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3 years ago

How many electrons will metals generally have in their outer shell?

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There is a maximum of two electrons in the outer shell.

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3 years ago

Read 2 more answers

The equilibrium of 2H 2 O(g) 2H 2 (g) + O 2 (g) at 2,000 K has a Keq value of 5.31 x 10-10. What is the Keq expression for this

nevsk [136]

Answer:

$5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

Explanation:

For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.

For a hypothetical reaction:

xA + yB ⇄ zC

The equilibrium constant is :

$Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }$

The given reaction involves the decomposition of H2O into H2 and O2

$2H_{2}O\rightleftharpoons 2H_{2} + O_{2}$

The equilibrium constant is expressed as :

$Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

Since Keq = 5.31*10^-10

$5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

3 0

3 years ago

Read 2 more answers

Indicate which solution in each pair has the lower pH. Your response should be a four letter "word". The first letter should be

JulijaS [17]

Answer:

bcfh

Explanation:

HClO₄ reacts with water thus:

HClO₄ + H₂O → H₃O⁺ + ClO₄⁻

That means HClO₄ produce H₃O⁺ that decreases pH. That means the higher concentration of HClO₄ decreases pH. Thus, lower pH will be:

b) 0.2 M HClO4

The reaction of NaClO₄ is:

NaClO₄ + H₂O → OH⁻ + HClO₄ + Na⁺

The higher concentration of NaClO₄ the higher production of OH⁻ that increase pH, that means the lower concentration of NaClO₄ the lower pH, thus, the answer is:

<em>c) 0.1 M NaClO or</em>

HF reacts with water thus;

HF ⇄ H⁺ + F⁻

The equilibrium constant is:

k = [H⁺] [F⁻] / [HF] = 3,5x10⁻⁴

For HNO₂ equilibrium is:

HNO₂ ⇄ H⁺ + NO₂⁻

k = [H⁺] [NO₂⁻] / [HNO₂] = 4,5x10⁻⁴

As k value is higher for HNO₂, the concentration of H⁺ will be higher in this system doing the HNO₂ with the lower pH.

f) 0.1 M HNO2

NaOH is a strong base that produce OH⁻ that increase pH, pure water is neutral, thus, the lowe pH is:

h) pure water

I hope it helps!

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4 years ago