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3 years ago

What are the differences between sand and potting soil? Are they both mixtures? How do you know?

Chemistry

1 answer:

katen-ka-za [31]3 years ago

7 0

What class is this for because it depends

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The molarity of a solution that contains 8.0 g of NaOH in a liter of solution is

Kay [80]

Answer:

0.2M NaOh

Explanation:

there are 0.2 mol of NaOH in 8.0 g. (8.0/40) =0.2. Molarity = mol/L = 0.2M.

8 0

2 years ago

Nitrosyl bromide decomposes according to the chemical equation below. 2NOBr(g) 2NO(g) + Br2(g) 1.00 atm of NOBr is sealed in a f

Andrew [12]

Given :

2NOBr(g) - -> 2NO(g) + Br2(g)

Initial pressure of NOBr , 1 atm .

At equilibrium, the partial pressure of NOBr is 0.82 atm.

To Find :

The equilibrium constant for the reaction .

Solution :

2NOBr(g) - -> 2NO(g) + Br2(g)

t=0 s 1 atm 0 0

$t=t_{eqb}$ 1( 1-2x) 2x x

So ,

$1-2x=0.82\\\\x=0.09$

At equilibrium :

$K_{eq}=\dfrac{[NO]^2[br_2]}{[NOBr]^2}\\\\K_{eq}=\dfrac{0.18^2\times 0.9}{0.82^2}\\\\K_{eq}=0.043\ atm$

Hence , this is the required solution .

3 0

3 years ago

Hey armies.. how are you doing

spayn [35]

Answer:

Nice and you

Explanation:

Please Mark me brainliest

3 0

2 years ago

Read 2 more answers

Consider the following reaction between mercury(II) chloride and oxalate ion.

Alina [70]

<u>Answer:</u> The rate law of the reaction is $\text{Rate}=k[HgCl_2][C_2O_4^{2-}]^2$

<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:

$2 HgCl_2(aq.)+C_2O_4^{2-}(aq.)\rightarrow 2Cl^-(aq.)+2CO_2(g)+Hg_2Cl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_4^{2-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_4^{2-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 2 from 1, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{(0.164)^a(0.45)^b}{(0.164)^a(0.15)^b}\\\\9=3^b\\b=2$

Dividing 2 from 3, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{(0.164)^a(0.45)^b}{(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_4^{2-}]^2$

3 0

3 years ago

The stronger the intermolecular forces, the _______the energy needed to separate the molecules, the ____________the boiling poin

slamgirl [31]

Answer:

higher, higher

Explanation:

It takes more energy to rip apart stronger bonds (that's mostly just common sense there). The boiling point increases because it would take more energy to get the molecules to go from a stuck together liquid, to separating in a gaseous form.

7 0

3 years ago