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lubasha [3.4K]
3 years ago
12

An example of a homogeneous mixture is 1. oil and vinegar dressing. 2. concrete. 3. a mixture of salt and pepper. 4. salt water

Chemistry
1 answer:
Sidana [21]3 years ago
4 0

A mixture in which composition is constant throughout the mixture is said to be homogeneous mixture.

Now, oil and vinegar dressing is not considered as a homogeneous mixture because composition is not uniform.

Concrete is made up of three components. Thus, composition is not uniform, so it is not a homogeneous mixture.

A mixture of salt and pepper, when salt is mixed in pepper then the mixture is not uniform as white flecks of salt and black flecks of pepper can be seen in the mixture. Thus, it is not a homogeneous mixture.

Salt water is uniform mixture. Thus,  it is a homogeneous mixture.

Thus, salt water is the correct answer.


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Lawrencium-262 has a half-life of 4 hr. How much of a 40 mg sample remains after 12 hours?
CaHeK987 [17]

Answer:

5 mg

Explanation:

If one half life is 4 hours, then 3 half lives is 12 hours.

This means that the sample will decay to 1/8 of its original amount.

So, the answer is 40(1/8) = 5 mg.

3 0
2 years ago
Please help!!
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Molarity =  Moles/Liter

Use the molecular atomic mass of NaCl to convert from grams to moles.
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so NaCl = 23 + 35.5 = 58.5 g/mol

multiply to cancel out grams
76 g NaCl * (1mol / 58.5 g NaCl) = 1.3 mol NaCl

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Which of these pathogens can be killed by antibiotics?
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5 0
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Jupiter has about 317 times more mass than Earth. What can you infer from this fact?
Ganezh [65]

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6 0
3 years ago
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The decomposition of N2O to N2 and O2 is a first-order reaction. At 730°C, the rate constant of the reaction is 1.94 × 10-4 min-
grin007 [14]

Answer:

Total pressure 5.875 atm

Explanation:

The equation for above decomposition is

2N_2O \rightarrow 2N_2 + O_2

rate constant k =  1.94\times 10^{-4} min^{-1}

Half life t_{1/2} = \frac{0.693}{k} = 3572 min

Initial pressure N_2 O = 4.70 atm

Pressure after 3572 min = P

According to first order kinematics

k = \frac{1}{t} ln\frac{4.70}{P}

1.94\times 10^{-4} = \frac{1}{3572} \frac{4.70}{P}

solving for P we get

P = 2.35 atm

2N_2O \rightarrow 2N_2 + O_2

initial           4.70                         0             0

change        -2x                          +2x           +x

final             4.70 -2x                     2x           x

pressure ofO_2 after first half life  = 2.35 = 4.70 - 2x

                                                          x = 1.175

pressure of N_2 after first half life  =  2x = 2(1.175) = 2.35 ATM

Total pressure  = 2.35 + 2.35 + 1.175

                          = 5.875 atm

5 0
3 years ago
Read 2 more answers
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