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pashok25 [27]
3 years ago
13

What is hte type of mixture whose components are evenly distrbuted out word wise?

Chemistry
1 answer:
Bumek [7]3 years ago
3 0
The answer is heterogeneous mixture.

Hope this helps
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What roles and responsibilities do the media have in reporting human right violation in a responsible manner
Flura [38]

You're question is similar to this one: brainly.com/question/3611487

which already has an answer.

6 0
3 years ago
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A 8.65-L container holds a mixture of two gases at 11 °C. The partial pressures of gas A and gas B, respectively, are 0.205 atm
Vsevolod [243]

 The  total pressure  = 1.402 atm


<u><em>calculation</em></u>

Total  pressure = partial  pressure  of gas A + partial pressure of gas B +  partial pressure  of third gas

partial  pressure  of gas A= 0.205 atm

Partial pressure of gas B =0.658 atm


partial pressure for third gas is calculated using ideal  gas equation

that is PV=nRT   where,

p(pressure)=? atm

V(volume) = 8.65 L

n(moles)= 0.200 moles

R(gas constant)=0.0821 L.atm/mol.k

T(temperature) = 11°c into kelvin =11+273 =284 k

make  p the subject of the formula by  diving both side by V

p =nRT/v


p = [(0.200 moles x 0.0821 L.atm/mol.K x 284 K)/8.65L)] =0.539 atm



Total  pressure  is therefore = 0.205 atm +0.658 atm +0.539 atm

=1.402 atm

6 0
3 years ago
what is the differential rate expression of the change in B with time? Incorporate elements that describe both its formation and
Ivanshal [37]
The differential rate expression for the rate of change in the concentration of B with time is
-rB = dCB/dt = kCB^n
where k is the rate constant and
n is the order of the reaction
This is assuming that the rate is only affected by the concentration of B and the order of the reaction is in the nth order.
5 0
3 years ago
What is bedrock layer​
Nuetrik [128]
Bed rock is a layer underneath minerals
6 0
3 years ago
Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO 3 ( s ) . The equation for the
Pavlova-9 [17]

Answer:

Mass of O_2 produced = 32 g

Explanation:

Calculation of the moles of KClO_3 as:-

Mass = 82.4 g

Molar mass of KClO_3 = 122.55 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{82.4\ g}{122.55\ g/mol}

Moles= 0.67237\ mol

From the reaction shown below:-

2KClO_3\rightarrow 2KCl+3O_2

2 moles of potassium chlorate on reaction forms 3 moles of oxygen gas

So,

0.67237 moles of potassium chlorate on reaction forms \frac{3}{2}\times  0.67237 moles of oxygen gas

Moles of oxygen gas = 1 mole

Molar mass of oxygen gas  = 32 g/mol

<u>Mass of O_2 produced = 32 g</u>

6 0
3 years ago
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