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3 years ago

How many oxigen are need it to produce two moleculas of water molecules

Chemistry

1 answer:

emmainna [20.7K]3 years ago

7 0

4 atoms of Oxygen are required to make 2 atoms of hydrogen. The compound H2O is 1 Hydrogen and 2 Oxygen.

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What happens when a catalyst is used in a chemical reaction

uranmaximum [27]

The reaction rate increases.

Why Well a catalyst usually lower the activation barrier in an energy diagram. The lower and smaller that gap means the reaction is taking place rapidly compared to when that activation barrier gap is higher.

6 0

3 years ago

How many grams of CO2 are produced from 6.7 L of O2 gas at STP?

Tasya [4]

<h3>Answer:</h3>

13 g CO₂

<h3>General Formulas and Concepts:</h3>

Chemistry

Atomic Structure

Reading a Periodic Table
Moles
STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Using Dimensional Analysis

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

Step 1: Define

Identify variables

[Given] 6.7 L O₂

[Solve] g O₂

Step 2: Identify Conversions

[STP] 22.4 L = 1 mol

[PT] Molar Mass of O: 16.00 g/mol

[PT] Molar Mass of C: 12.01 g/mol

Molar Mass of CO₂: 12.01 + 2(16.00) = 44.01 g/mol

Step 3: Convert

[DA] Set up: $\displaystyle 6.7 \ L \ O_2(\frac{1 \ mol \ O_2}{22.4 \ L \ O_2})(\frac{44.01 \ g \ O_2}{1 \ mol \ O_2})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 13.1637 \ g \ O_2$

Step 4: Check

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

13.1637 g CO₂ ≈ 13 g CO₂

5 0

3 years ago

Based on the information that is given, which atom is the table ?

andriy [413]

Answer:

The one with the greatest mass would be the one that has the most things in the nucleus, protons and nutrons

Explanation:

3 0

3 years ago

Ethylene glycol flows at 0.01 kg/s through a 3-mm diameter, thin-walled tube. The tube is coiled and submerged in well-stirred w

leonid [27]

Answer:

heat rate= 1281W

length = 15.8m

Explanation:

we have this data to answer this question with

Tmi = 85 degrees

Tmo = 35 degrees

Ts = 25 dgrees

flow rate = 25 degrees

using engine oil property from table a-5

Tm = Tmo - TMi/2 = 333k

u =0.522x10⁻²

k = 0.26

pr = 51.3

cp = 2562 J/kg.k

mcp(Tmo-Tmi) =

0.01 x 2562(35-85)

= 1281 W

we find the change in Tim

= [(35-25)-(85-25)]/ln[(35-25)/(85-25)]

= -50/ln0.167

= -50/-1.78976

= 27.9°c

we finf the required reynold number

4x0.01/πx0.003x0.522x10⁻²

= 0.04/0.00004921

= 812.8

= 813

we find approximate correlation

NuD = hd/k

NuD = 3.66

3.66 = 0.003D/0.26

cross multiply

0.003D = 3.66x0.26

D = 3.66x0.26/0.003

= 317.2

As = 1281/317x27.9

= 0.145

As = πDL

L = As/πD

= 0.145/π0.003

= 0.145/0.009429

L = 15.378

4 0

3 years ago

Part a how many grams of xef6 are required to react with 0.579 l of hydrogen gas at 6.46 atm and 45°c in the reaction shown belo

Nina [5.8K]

First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.

PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂

The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = 11.69 g

6 0

4 years ago