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2 years ago

Based on the periodic table, which of the following elements has properties most like neon’s?

Chemistry

2 answers:

Sonbull [250]2 years ago

7 0

D.krypton (Kr).............

ohaa [14]2 years ago

3 0

Answer:

<h2>D. Krypton (Kr)</h2>

Explanation:

because

Neon and Krypton, both are belongs to Inert or Noble gas Group

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Does a spring tide always happen in the season of spring

grin007 [14]

THE SPRING TIDE DOES NOT HAVE ANYTHING TO DO WITH THE SEASON OF SPRING AND IT OCCURS MOSTLY IN THE FALL

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2 years ago

How do the process of crabon cycle contribute to climate change.

bonufazy [111]

Explanation:

Carbon that is a part of rocks and fossil fuels like oil, coal, and natural gas may be held away from the rest of the carbon cycle for a long time. These changes add more greenhouse gases in our atmosphere and this causes climate change.

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3 years ago

A 150.0 mL sample of a 1.50 M solution of CuSO4 is mixed with a 150.0 mL sample of 3.00 M KOH in a coffee cup calorimeter. The t

svp [43]

Mols CuSO4 = M x L = 1.50 x 0.150 = 0.225
mols KOH = 3.00 x 0.150 = 0.450 
specific heat solns = specific heat H2O = 4.18 J/K*C 

CuSO4 + 2KOH = Cu(OH)2 + 2H2O 
q = mass solutions x specific heat solns x (Tfinal-Tinitial) + Ccal*deltat T 
q = 300g x 4.18 x (31.3-25.2) + 24.2*(31.3-25.2) 
dHrxn in J/mol= q/0.225 mol CuSO4 
Then convert to kJ/mol

5 0

3 years ago

Read 2 more answers

11C. Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solut

Llana [10]

For the reactants,

The oxidation number of hydrogen = +1
The oxidation number of oxygen = -2
The oxidation number of arsenic = +5
The oxidation number of carbon = +3

For the products,

The oxidation number of hydrogen = +1
The oxidation number of oxygen = -2
The oxidation number of arsenic = +3
The oxidation number of carbon = +4

Here, arsenic (+5 to +3) and carbon (+3 to +4) are the only oxidation numbers changing.

Note that an increase in oxidation number means electrons are lost. Thus oxidation is occurring, and a decrease in oxidation number means electrons are being gained, and thus reduction is occurring.

Also, the compound that contains the element being oxidized is the reducing agent, and the compound that contains the element being reduced is the oxidizing agent.

So, the answers are:

name of the element oxidized: Carbon

name of the element reduced: Arsenic

formula of the oxidizing agent: $\text{H}_{3}\text{AsO}_{4}$

formula of the reducing agent: $\text{H}_{2}\text{C}_{2}\text{O}_{4}$

6 0

1 year ago

g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete

Mandarinka [93]

Answer:

The concentration of the sodium and arsenate ions at the end of the reaction in the final solution

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Explanation:

Complete Question

A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)

Concentration in mol/L = (Number of moles) ÷ (Volume in L)

Number of moles = (Concentration in mol/L) × (Number of moles)

For Na₂HAsO₄

Concentration in mol/L = 0.03798 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03798 × 0.5 = 0.01899 mole

For NaOH

Concentration in mol/L = 0.03428 M

Volume in L = (500/1000) = 0.50 L

Number of moles = 0.03428 × 0.5 = 0.01714 mole

Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.

Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O

0.01899 0.01714 0 0 (At time t=0)

(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)

0.00185 | 0 → 0.01714 0.01714 (end)

Hence, at the end of the reaction, the following compounds have the following number of moles

Na₂HAsO₄ = 0.00185 mole

This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction

NaOH = 0 mole

Na₃AsO₄ = 0.01714 moles

This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction

H₂O = 0.01714 moles

So, at the end of the reaction

Na⁺ has 0.0037 + 0.05142 = 0.05512 mole

(HAsO₄)²⁻ has 0.00185 mole

(AsO₄)³⁻ has 0.01714 mole.

And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L

Hence, the concentration of the sodium and arsenate ions at the end of the reaction is

[Na⁺] = 0.05512 M

[HAsO₄²⁻] = 0.00185 M

[AsO₄³⁻] = 0.01714 M

Hope this Helps!!!

8 0

3 years ago

Read 2 more answers