Chem help please. i cant figure out this problem

Chemistry

1 answer:

xxTIMURxx [149]3 years ago

3 0

11 a- ions would exist in the solution, they can carry charges.

B-solids that are neutral, they wouldn't conduct much of a current.

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(BRAINLIEST FOR CORRECT AND BEST EXPLAINED ANSWER) What properties distinguish solids, liquids, gases, and plasma from each othe

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Explain in terms of atomic structure why the atomic radius of K is larger than that of Na

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Potassium (K) has a larger average atomic radius (220 pm) than sodium (Na) does (180 pm). The potassium atom has an extra electron shell compared to the sodium atom, which means its valence electrons are further from the nucleus, giving potassium a larger atomic radius. The ionic radius increases in a particular group on moving from top to bottom due to increase in the principle energy shell though the number or electrons in the valence shell remain the same

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3 years ago

What mass of silver oxide, ag2o, is required to produce 25.0 g of silver sulfadiazine, agc10h9n4so2, from the reaction of silver

LenaWriter [7]

Step one write the chemical equation for reaction
= Ag2O + 2(C10H10N4SO2)---> 2 ( AgC10H9N4SO2)

The reacting ratio of Ag2O to AgC10H9N4SO2 is 1:2 from the reaction above

step 2; find the number of moles of AgC10H9N4SO2
that is mass/molar mass
The molar mass of AgC10H9N4SO2 = 107.86 +(12 x10) + (1 x 9) + (4 x 14) +32 +(16 x2) =356.86g/mol
moles is therefore= 25g/356.86g/mol= 0.07moles

by use of mole ratio the moles of Ag2O= 0.0702=0.035moles
mass = moles x molar mass
the molar mass of Ag2O=231.72 g/mol
mass is therefore= 231.72g/mol x 0.035moles= 8.11grams

5 0

4 years ago

How many grams of CO2 will be produced when 8.50 g of methane react with 15.9 g of O2, according to the following reaction? CH4(

Vedmedyk [2.9K]

Taking into account the reaction stoichiometry, 10.93 grams of CO₂ are formed when 8.50 g of methane react with 15.9 g of O₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CH₄: 1 mole
O₂: 2 moles
CO₂: 1 mole
H₂O: 2 moles

The molar mass of the compounds is:

CH₄: 16 g/mole
O₂: 32 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CH₄: 1 mole ×16 g/mole= 16 grams
O₂: 2 moles ×32 g/mole= 64 grams
CO₂: 1 mole ×44 g/mole= 44 grams
H₂O: 2 moles ×18 g/mole=36 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 16 grams of CH₄ reacts with 64 grams of O₂, 8.50 grams of CH₄ reacts with how much mass of O₂?

$mass of O_{2} =\frac{8.50 grams of CH_{4}x64 grams of O_{2} }{16grams of CH_{4}}$

mass of O₂= 34 grams

But 34 grams of O₂ are not available, 15.9 grams are available. Since you have less mass than you need to react with 8.50 grams of CH₄, O₂ will be the limiting reagent.

<h3>Mass of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 64 grams of O₂ form 44 grams of CO₂, 15.9 grams of O₂ form how much mass of CO₂?

$mass of CO_{2} =\frac{15.9 grams of O_{2}x44 grams of CO_{2} }{64grams of O_{2}}$