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kobusy [5.1K]
4 years ago
15

Is every compound that contains hydrogen an acid?

Chemistry
1 answer:
Vikki [24]4 years ago
3 0
No every compound that contains isnt an acid .
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Ksivusya [100]

Answer:

A, C, D.

Explanation:

6 0
3 years ago
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How many moles of oxygen must be placed
umka21 [38]

Answer: 0.245 moles of oxygen must be placed in the container to exert the given pressure at the given temperature. The Ideal Gas Law equation gives the relationship among the pressure, volume, temperature, and moles of gas.

Further Explanation:

The Ideal Gas Equation is:  

PV = nRT  

where:  

P - pressure (in atm)  

V - volume (in L)  

n - amount of gas (in moles)  

R - universal gas constant 0.08206 \frac{L-atm}{mol-K}  

T - temperature (in K)  

In the problem, we are given the values:  

P = 2.00 atm (3 significant figures)

V = 3.00 L  (3 significant figures)

n = ?

T = 25.0 degrees Celsius (3 significant figures)  

We need to convert the temperature to Kelvin before we can use the Ideal Gas Equation. The formula to convert from degree Celsius to Kelvin is:  

Temperature \ in \ Kelvin = Temperature\ in \ Celsius \ + \ 273.15  

Therefore, for this problem,  

Temperature\ in \ K = 25.0 +273.15\\Temperature\ in \ K = 298.15  

Solving for n using the Ideal Gas Equation:  

n \ = \frac{PV}{RT}\\n \ = \frac{(2.00 \ atm) \ (3.00 \ L)}{(0.08206 \ \frac{L-atm}{mol-K})( 298.15 \ K)}  \\n \ = 0.245 \ mol

The least number of significant figures is 3, therefore, the final answer must have only 3 significant figures.

Learn More  

1. Learn more about Boyle's Law brainly.com/question/1437490  

2. Learn more about Charles' Law brainly.com/question/1421697  

3. Learn more about Gay-Lussac's Law brainly.com/question/6534668

Keywords: Ideal Gas Law, Volume, Pressure

4 0
4 years ago
What does energy mean?
Fofino [41]

Answer:

strength and vitality required for sustained physical or mental activity

Explanation:

7 0
3 years ago
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How many molecules are in 1kg of water
Mila [183]

Answer:

334.2× 10²³ molecules

Explanation:

Given data:

Mass of water = 1 Kg ( 1000 g )

Number of molecules = ?

Solution:

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 1000 g/ 18 g/mol

Number of moles = 55.5 mol

1 mole contain 6.022× 10²³ molecules

55.5 mol×6.022× 10²³ molecules

334.2× 10²³ molecules

8 0
3 years ago
slader Consider the following reactions: A. uranium-238 emits an alpha particle; B. plutonium-239 emits an alpha parti- cle; C.
pishuonlain [190]

<u>Answer:</u>

<u>For A:</u> The equation is _{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha

<u>For B:</u> The equation is _{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha

<u>For C:</u> The equation is _{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta

<u>Explanation:</u>

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

<u>For A:</u> Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha

<u>For B:</u> Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha

<u>For C:</u> Thorium-239 emits a beta particle

The nuclear equation for this process follows:

_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta

6 0
4 years ago
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