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Ronch [10]
3 years ago
10

How to calculate the mass percent of hydrogen

Chemistry
1 answer:
Keith_Richards [23]3 years ago
8 0

Answer:

divide the mass of element in 1 mole of the compound by the compound's molar mass and multiply the answer by 100.

Explanation:

take the molar mass of hydrogen in the water molecule, divide by the total molar mass of water, and multiply by 100.

does it help?

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at stp which of following would have the same number of molecules a 1 l of c2h4 gas? a. 0.5 of H2 b. 1L of Ne c.2L of H2O d.3L o
sashaice [31]

Answer:d

Explanation:

4 0
3 years ago
A certain gas at 2oC and 1.00 atm pressure fills a 4.0 container. What volume will the gas occupy at 100oC and 780 torr pressure
artcher [175]

The final volume V₂=4.962 L

<h3>Further explanation</h3>

Given

T₁=20 + 273 = 293 K

P₁= 1 atm

V₁ = 4 L

T₂=100+273 = 373 K

P₂=780 torr=1,02632 atm

Required

The final volume

Solution

Combined gas law :

P₁V₁/T₁=P₂V₂/T₂

Input the value :

V₂=(P₁V₁T₂)/(P₂T₁)

V₂=(1 x 4 x 373)/(1.02632 x 293)

V₂=4.962 L

6 0
3 years ago
If you have 2.1 liters of hydrogen gas (at STP), what mass of hydrogen gas would you have?
GuDViN [60]

Answer:

0.19 g

Explanation:

Step 1: Given data

Volume of hydrogen at standard temperature and pressure (STP): 2.1 L

Step 2: Calculate the moles corresponding to 2.1 L of hydrogen  at STP

At STP (273.15 K and 1 atm), 1 mole of hydrogen has a volume of 22.4 L if we treat it as an ideal gas.

2.1 L × 1 mol/22.4 L = 0.094 mol

Step 3: Calculate the mass corresponding to 0.094 moles of hydrogen

The molar mass of hydrogen is 2.02 g/mol.

0.094 mol × 2.02 g/mol = 0.19 g

6 0
3 years ago
Can someone pls help, I’m sorry if it’s a lot
Helga [31]
Answer: 2738


Explaining: you gotta add
3 0
3 years ago
The spectrochemical series is I &lt; Br&lt; &lt; Cl^- &lt; F
vfiekz [6]

Answer:

b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

Explanation:

The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.

Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.

Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.

Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.

6 0
3 years ago
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