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3 years ago

9

Sodium chlorate decomposes into sodium chloride and oxygen gas as seen in the equation below.

2 answers:

Wewaii [24]3 years ago

8 0

Answer:

10 moles of sodium chloride

Explanation:

$\sf{}$

Andru [333]3 years ago

3 0

Answer:

${ \rm{2\: moles \: of \: sodium \: chlorate = 2 \: moles \: of \: sodium \: choride}} \\ { \rm{10 \: moles \: of \: sodium \: chlorate = ( \frac{10}{2} \times 2) \: moles}} \\ \\ = { \boxed{ \rm{ \: 10 \: moles \: of \: sodium \: chloride}}} \\$

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what is the suggested weight of the sample that should be disssolved in the beaker before it is transferred to the NMR tube

sukhopar [10]

The sample must be sufficiently soluble (fig. 2) to yield an NMR spectrum. For 1H and 1H observed NMR, it is recommended to dissolve between 2 and 10 mg in between 0.6 and 1 mL of solvent so that the sample depth is at least 4.5 cm in the tube (fig. 3).

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2 years ago

The equilibrium constant for the reaction NO2(g)+NO3(g)→N2O5(g) is 2.1x10-20 , therefore: a. At equilibrium, the concentration

frutty [35]

Answer: c. At equilibrium, the concentration of reactants is greater than the products

Explanation:

Equilibrium constant for a reaction is the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

For the reaction:

$NO_2(g)+NO_3(g)\rightleftharpoons N_2O_5(g)$

Equilibrium constant is given as:

$K_{eq}=\frac{[N_2O_5]}{[NO_2]\times [NO_3]}$

$2.1\times 10^{-20}=\frac{[N_2O_5]}{[NO_2]\times [NO_3]}$

When

a) K > 1, the concentration of products is greater than the concentration of reactants

b) K < 1, the concentration of reactants is greater than the concentration of products

c) K= 1, the reaction is at equilibrium, the concentration of reactants is equal to the concentration of products

Thus as $K_{eq}$ is $2.1\times 10^{-20}$ which is less than 1,

the concentration of reactants is greater than the concentration of products

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4 years ago

Are the phosphate groups on the interior or the exterior of the DNA molecule?

Makovka662 [10]

the double helix is hydrogen bonded through the bases only so the bases are inside the helix only

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3 years ago

WILL MAKE BRAINEST

Elanso [62]

V=84.0 mL = 84.0 cm³
m=609.0 g

p=m/v

p=609.0/84.0=7.25 g/cm³

8 0

3 years ago

At a certain temperature, 0.900 mol SO 3 is placed in a 2.00 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equil

puteri [66]

Answer: The value of $K_c$ is 0.0057

Explanation:

Initial moles of $SO_3$ = 0.900 mole

Volume of container = 2.00 L

Initial concentration of $SO_3=\frac{moles}{volume}=\frac{0.900moles}{2.00L}=0.450M$

equilibrium concentration of $O_2=\frac{moles}{volume}=\frac{0.110mole}{2.00L}=0.055M$ [/tex]

The given balanced equilibrium reaction is,

$2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)$

Initial conc. 0.450 M 0 0

At eqm. conc. (0.450 -2x) M (2x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[O_2][SO_2]^2}{[SO_3]^2}$

$K_c=\frac{x\times (2x)^2}{0.450-2x)^2}$

we are given : x = 0.055

Now put all the given values in this expression, we get :

$K_c=\frac{0.055\times (2\times 0.055)^2}{0.450-2\times 0.055)^2}$

$K_c=0.0057$

Thus the value of the equilibrium constant is 0.0057

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4 years ago