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Diano4ka-milaya [45]
3 years ago
12

What kind of mixture is a solution

Chemistry
2 answers:
aleksandrvk [35]3 years ago
8 0
A simple solution is basically two substances<span> that are evenly mixed together. One of them is called the solute and the other is the solvent. A solute is the substance to be dissolved (sugar). The solvent is the one doing the dissolving (water).</span>
blagie [28]3 years ago
3 0
<span>The answer is homogeneous.

</span><span>The best description is homogeneous. The particles which are dissolved are invisible, which is why solutions, unlike colloids or suspensions, are always clear.</span>
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Identify and calculate the number of representative particles in 2.15 moles of gold.<br>​
LekaFEV [45]

Answer:

1.29 * 10^{24} particles of gold

Explanation:

To convert the number of moles of any substance, in this case gold, you need Avogadro's number.

Avogadro's number is always 6.022 × 10^{23}

2.15 moles Au × \frac{6.022*10^{23} particles}{1 mole Au} = 1.29 * 10^{24} particles of gold

5 0
3 years ago
The ph of a 0.30 m solution of a weak acid is 2.67. what is the ka for this acid?
azamat
To determine the Ka of the acid, we can use the equation for the pH of weak acids which is expressed as:

pH = -0.5 log Ka
2.67 = -0.5 log Ka
Ka = 4.571x10^-6

Weak acids are acids that do not dissociate completely in solution. The solution would contain the cations, anions and the acid itself as a compound. Hope this helps.
7 0
3 years ago
Read 2 more answers
A sample of pure NO2 is heated to 335 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+
vodka [1.7K]

The equilibrium constant for the reaction is 0.00662

Explanation:

The balanced chemical equation is :

2NO2(g)⇌2NO(g)+O2(g

At t=t  1-2x ⇔ 2x + x moles

The ideal gas law equation will be used here

PV=nRT

here n= \frac{w}{W} = \frac{w}{V}= density

P = \frac{density RT}{M}       density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm

putting the values in reaction

0.75 = \frac{0.525 x 0.0821 x 608.15 }{M}

  M    = 34.61

         

to calculate the Kc

Kc=\frac{ [NO] [O2]}{NO2}

  \frac{1-2x}{1+x} x M NO2 + \frac{2x}{1+x} M NO+ \frac{x}{1+x} M O2

Putting the values as molecular weight of NO2, NO,O2

\frac{46(1-2x) +30(2x)+32x}{1+x}

34.61= \frac{46}{1+x}

x= 0.33

Kc= \frac{4x^2)x}{1-2x^2}

    putting the values in the above equation

Kc = 0.00662

     

5 0
3 years ago
ML? PLEASEEE HELP<br> What is the answer
Fofino [41]
22.5 milliliters :)))
6 0
3 years ago
Read 2 more answers
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
Nastasia [14]

Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

5 0
3 years ago
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