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LekaFEV [45]
2 years ago
13

Activity/Exercises: Kobe has been living in a coastal barangay for 30 years. Recently, he noticed that during high tides, water

could reach high altitude as compared five years ago. The water could go further inland. What happened to the coasts near his barangay that enabled water to reach further inland? Write your answer below:​
Chemistry
1 answer:
Akimi4 [234]2 years ago
6 0
  • The sea water during high tides enters inland.
  • It harms the soul.
  • It makes the soul salty.
  • which makes hard to survive for a plant.
  • It harms the atmosphere too
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Using the trend for chlorine, create a rule for naming halogen oxyacids.
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All oxayacids have cations so no need to name the cation (H+) If name of polyatomic anion ends in -ate change to -ic for acid and if it ends with -ite change to -ous for acid

example:

ion nitrate is called nitric acid
ion nitrate is called nitrous acid
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Un ácido nítrico tiene una densidad de 10% ,calcular el peso de 1000 centímetros cúbicos
devlian [24]
D = M/V
D = 0.10
V = 1000 cm^3

0.10 = M / 1000
M = 100

Porque la densidad es 10%, se puede usar 0.10 en este formulario para calcular el peso. No sé lo que es la unidad para el peso pero es 100.
4 0
3 years ago
An ideal gas is a gas. <br> Perfect<br> Theoretical<br> Real
dusya [7]

Answer:

An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions. The ideal gas concept is useful because it obeys the ideal gas law, a simplified equation of state, and is amenable to analysis under statistical mechanics.

5 0
3 years ago
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___CH4 + ___O2 → ___CO2 + ___H2O
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CH4 + 2O2 =======> CO2 + 2H2O
5 0
3 years ago
If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W
olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3

Therefore, the leftover of sodium nitrate is:

m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3

Finally, the percent yield is computed via:

Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%

Best regards!

6 0
3 years ago
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