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Lunna [17]
3 years ago
6

How to separate water from a mixture of rice and peas? Please

Chemistry
1 answer:
svp [43]3 years ago
8 0
Depending on the type of pea, the peas should come to the top as they are lighter and larger. Then just scoop or brush them off into another container. If you separate it into smaller batches and work them one by one, you should then find you get pure rice batches left which you can put in their own container.
You might be interested in
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

8 0
3 years ago
Calculate a reasonable amount (mass in g) of your unknown acid to use for a titration. You will want about 30 mL of titrant to g
Vinil7 [7]

Answer:

"0.60 g" is the appropriate solution.

Explanation:

The given values are:

Volume of base,

= 30 ml

Molarity of base,

= 0.05 m

Molar mass of acid,

= 400 g/mol

As we know,

⇒  Molarity=\frac{Number \ of \ moles \ of \ base}{Number \ of \ solution}

On substituting the values, we get

⇒           0.05=\frac{Number \ of \ moles \ of \ base}{30\times 10^{-3}}

⇒  Number \ of \ moles \ of \ base=0.05\times 30\times 10^{-3}

⇒                                             =1.5\times 10^{-3}  

hence,

⇒  Moles \ of \ acid=\frac{Mass \ of \ acid}{Molar \ mass \ of \ acid}

On substituting the values, we get

⇒  1.5\times 10^{-3}=\frac{Mass \ of \ acid}{400}

⇒  Mass \ of \ acid=1.5\times 10^{-3}\times 400

⇒                         =0.60 \ g

8 0
3 years ago
As you go down group one of the periodic table, the reactions become and more
Lady bird [3.3K]

Answer:

vigorous

Explanation:

As you go down group one of the periodic table, the reactions become and more vigorous.

3 0
2 years ago
Which metalloid is often found in microchips, semiconductors, and sand?a. boron b. arsenic c. germanium d. silicon​
elena-14-01-66 [18.8K]

Answer:

silicon

Explanation:

Silicon. Silicon is often found in electronic devices, as well as the sand.

hope this helps UwU

5 0
2 years ago
Read 2 more answers
Under identical conditions of temperature and pressure, which of the following gasses is the densest: Ne, CO2, or Cl2?
Tanzania [10]
Cl2=3.17g/L
Ne=.901g/L
CO2=1.96g/l
 therefore Cl2 is the densest gas under the given conditions.
7 0
3 years ago
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