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fgiga [73]
3 years ago
15

A solution that contains a large amount of salt and a small amount of water is said to be a _______ solution. A. diffused B. uns

aturated C. dilute D. concentrated
Chemistry
2 answers:
stepladder [879]3 years ago
7 0
The correct answer for the blank is “concentrated.”
Pachacha [2.7K]3 years ago
5 0

Answer:

D. Concentrated

Explanation:

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Uestion 3 (1 point)
elena-s [515]

Answer:

density=mass/volume.

density=14/18

=0.777777777 ~ 0.8g/ml.

It will float because the it lesser than the density of water.

6 0
3 years ago
Read 2 more answers
What is the ROUNDED atomic number for this elements?
denis-greek [22]
Sodium/Atomic number

11

Gold/Atomic number

79
Potassium/Atomic number

19

Silicon/Atomic number

14




8 0
3 years ago
The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10
Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

                   2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0        125 kPa

At t = teq     125 - 2x      4x        x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 2.8\times 10^{-3} min⁻¹

Initial concentration [A_0] = 125 kPa

Final concentration [A_t] = 91 kPa

Time = ?

Applying in the above equation, we get that:-

91=125e^{-2.8\times 10^{-3}\times t}

125e^{-2.8\times \:10^{-3}t}=91

-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)

t=113\ min

3 0
3 years ago
Which of the following is created when solids, liquids, or gases mix together with one another?
gavmur [86]

the correct answer is option ( A ) Mixture

Explanation:

It is defined as the substance that is made by the combination of two or more different components. Or, we can say that it is created when solids liquids or gases mixed with one another. Hence, the correct option is, mixture.

Hope it helps you

Mark my answer as brainlist

have a nice day

8 0
3 years ago
What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.
Hitman42 [59]

Explanation:

The given data is as follows.

      [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,

         V = 500 ml,   [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

   No. of moles of benzoic acid = Molarity × Volume

                         = 2 \times 0.475

                         = 0.095 mol

And, moles of NaOH present in the solution will be as follows.

    No. of moles of NaOH = Molarity × Volume

                          = 2 \times 0.025

                          = 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

         C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O

Initial:        0.095           0.05            0             0

Equlbm:  (0.095 - 0.05)  0            0.05

        pH = pK_{a} + log \frac{Base}{Acid}  

              = 4.2 + log \frac{0.05}{0.045}

              = 4.245

For,  

         HCOOH + NaOH \rightarrow HCOONa + H_{2}O

Initial:       0.2x     2(0.5 - x)               0

Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)

As,

           pH = pK_{a} + log \frac{Base}{Acid}  

          4.245 = 3.75 + log \frac{Base}{Acid}

      log \frac{Base}{Acid} = 0.5

    \frac{Base}{Acid} = 3.162

Now,

        \frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 3.162

               x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

                             = 0.036 L

                             = 36 ml               (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

                 

8 0
3 years ago
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