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elixir [45]
2 years ago
10

Which quantity represents 0.500 Mole at STP

Chemistry
1 answer:
KIM [24]2 years ago
4 0
0.500 moles is roughly .5*6.022*10^23=3.011*10^23 atoms. This is independent of STP.
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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘c is 2000 m/s. Note that 1. 0 mol of diatomic hydrogen at
Leno4ka [110]

The rms speed will be 500 m/s

<h3>What is Root mean square speed ?</h3>

Root mean square speed (Vrms) is defined as the square root of the mean of the square of speeds of all molecules.

Root mean square speed (vrms) Root mean square speed (vrms) is defined as the square root of the mean of the square of speeds of all molecules

It is given that

Speed of a diatomic hydrogen molecule,2000 m/s

Mol of diatomic hydrogen,1.0

Temperature,50°C

The rms speed of diatomic molecule will be:

√(3KT)/( m)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

(3/2)KT = (1/2) mv²

v = √(3KT)/m  

FOR H₂:  v = √(3KT)/m = 2000 m/s  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

Therefore the  rms (root-mean-square) speed of a oxygen molecule at 50∘c is 500m/s.

To know more about Root mean square speed

brainly.com/question/7213287

#SPJ4

3 0
2 years ago
One part of the cell theory states that all living things are made up of one or more cells. Scientists had to find ways to test
sergejj [24]

Answer:

A population or community research line can be carried out, wherever at a certain point in time, regardless of whether it is a cross-sectional study.

In addition, the people who would be the population to be studied or the object of study might or might not know the cause of the study (blind) while the researcher could be experimentally participatory.

Explanation:

They are prevalence studies, in which the presence of a health condition or state is determined in a well-defined population and in a determined time frame: one day, one week, a particular moment in life, even if it does not temporarily coincide in all the subjects (for example, the blood pressure figures at the time of entering the school or at the beginning of the holidays, the prevalence of diabetes in hospitalized patients on a given day, etc.).

They are like "photographs" of a state of affairs at a given moment. The simultaneous determination of what is understood by exposure and event does not allow defining causality.

5 0
2 years ago
Part d based on your answer to part c, how many bonding electrons and lone pair electrons (nonbonding electrons) are there in th
elena55 [62]
The Lewis Structure of HCN is shown below,

Number of Bonding Electrons:
                                               In HCN Hydrogen is bonded to Carbon through single bond and Nitrogen is bonded to Carbon through Triple Bond. Single bond is formed by two bonding electrons, while, triple bond is formed by six bonding electrons, Hence,

                               Number of Bonding Electrons  =  8

Number of Non-Bonding Electrons:
                                                            In HCN there is only one lone pair of electron present on Nitrogen atom which is not taking part in bonding. Hence, 
                              Number of Non-Bonding Electrons  =  2

Result:

                               Number of Bonding Electrons  =  8

                              Number of Non-Bonding Electrons  =  2

3 0
3 years ago
Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

6 0
3 years ago
Which of the following is not a correct chemical equation for a double displacement reaction?
sergeinik [125]
Where are the options?
6 0
3 years ago
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