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3 years ago

Which scientist was the first to practice the scientific method we use today

Chemistry

1 answer:

bixtya [17]3 years ago

6 0

Answer:

Aristotle

Explanation:

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2H2 + O2 --> 2H2O How many grams of oxygen gas are required to produce 90. g of water?

Travka [436]

Answer:

80g

Explanation:

2H2 + O2 —> 2H2O

MM of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass conc. of H2O from the balanced equation = 2 x 18 = 36g

MM of O2 = 16 x 2 = 32g/mol

From the equation,

32g of O2 reacted to produce 36g of H2O.

Therefore Xg of O2 will react to produce 90g of H2O i.e

Xg of O2 = (32x90)/36 = 80g

8 0

3 years ago

Help please ASAP!!!

vodka [1.7K]

Silicon has 14 protons and it should have at least 12 neutrons

8 0

3 years ago

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I need answers plz btw D says ice crystals

tangare [24]

Answer:

B. Hydrogen and oxygen molecules

General Formulas and Concepts:

Chemistry - Reactions

Reactants are always on the left side of the arrow
Products are always on the right side of the arrow

Explanation:

Step 1: Define

Reaction RxN: 2H₂ + O₂ → 2H₂O

Step 2: Identify

Reactants: H₂ and O₂

Products: H₂O

7 0

3 years ago

`I am holding a balloon containing 439 mL of gas over my fireplace. The temperature and pressure of the gas inside the balloon i

Andrew [12]

Answer:

- 0.07 °C

Explanation:

At constant pressure and number of moles, Using Charle's law

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

Given ,

V₁ = 439 mL = 0.439 L ( 1 L = 0.001 mL )

V₂ = 0.378 L

T₁ = 317.15 K

T₂ = ?

Using above equation as:

$\frac{0.439}{317.15}=\frac{0.378}{T_2}$

$T_2=\frac{0.378\cdot \:317.15}{0.439}=273.08\ K$

The conversion of T(K) to T( °C) is shown below:

T( °C) = T(K) - 273.15

So, T = 273.08 - 273.15 °C = - 0.07 °C

3 0

4 years ago

How many molecules of oxygen are there in 1.00 dm3 of oxygen at standard pressure and a temperature of 273 K. Give your answer t

rewona [7]

Pressure=p=1atm
Temperature=T=273K.
V=1dm^3=1L

we need no of moles

$\\ \tt\hookrightarrow PV=nRT$

$\\ \tt\hookrightarrow n=\dfrac{PV}{RT}$

$\\ \tt\hookrightarrow n=\dfrac{1}{273(8.314)}$

$\\ \tt\hookrightarrow n=1/2269.7=0.0004mol$

No of molecules=No of moles×Avagadro no

$\\ \tt\hookrightarrow 0.0004(6.023\times 10^{23})$

$\\ \tt\hookrightarrow 2.4\times 10^{20} molecules$

5 0

2 years ago

Read 2 more answers