Which has a larger radius Me or Xe?

WILL GIVE BRAINLIEST, Which of the following would have the greatest amount of surface area?

Elodia [21]

Answer:

C. 1 cubic foot of loose sand

Explanation:

For many objects having equal volume , surface area will be maximum

of the object which has spherical shape .

But when a sphere is broken into tiny small spheres , total surface area of all the small spheres will be more than surface area of big sphere .

Hence among the given option , surface area of loose sand will have greatest surface area . Loose sand is equivalent to small spheres .

5 0

3 years ago

KOH (aq) + HNO(little3) (aq) = KNO(little 3) (aq) + H20 (1)

Yuki888 [10]

The reaction already balanced

<h3>Further explanation </h3>

Equalization of chemical reactions can be done using variables. Steps in equalizing the reaction equation:

• 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c, etc.

• 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index (subscript) between reactant and product

• 3. Select the coefficient of the substance with the most complex chemical formula equal to 1

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C, H and the last O

Reaction

KOH (aq) + HNO₃ (aq) ⇒ KNO₃ (aq) + H₂0 (l)

give coefficient

aKOH (aq) + bHNO₃ (aq) ⇒ KNO₃ (aq) + cH₂0 (l)

K : left = a, right = 1⇒a=1

N : left = b, right =1, b=1

H: left = a, right a+b=2c⇒1+1=2c⇒2=2c⇒c=1

KOH (aq) + HNO₃ (aq) ⇒ KNO₃ (aq) + H₂0 (l)⇒ already balanced

5 0

3 years ago

Which of the following is the balanced equation for this process?

Hunter-Best [27]

The answer is A.Hope this helps

8 0

3 years ago

A 25.0 ml sample of 0.150 m benzoic acid is titrated with a 0.150 m naoh solution. what is the ph at the equivalence point? the

Ad libitum [116K]

Solution:

At the equivalence point, moles NaOH = moles benzoic acid

HA + NaOH ==> NaA + H2O where HA is benzoic acid

At the equivalence point, all the benzoic acid ==> sodium benzoate

A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)

Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11

Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150

x^2 = 3.33x10^-12

x = 1.8x10^-6 = [OH-]

pOH = -log [OH-] = 5.74

pH = 14 - pOH = 8.26

5 0

3 years ago

Read 2 more answers

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$