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defon
3 years ago
10

Formula for Sodium Hydrogen Carbonate NaHCO3? or Na2HCO3?

Chemistry
1 answer:
yan [13]3 years ago
8 0

NaHCO3 is the right answer

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Ryan and Abigail are conducting an experiment. They place six ice cubes into a beaker and find the combined mass. They allow the
JulijaS [17]

The mass of the ice cubes and the water will be equal because the same amount of matter is in the beaker.

Matter is anything that has mass and occupy space. All substances are composed of matter. According to the law of conservation of mass, matter can neither be created nor destroyed but can be converted from one form to another.

Since mass is the quantity of matter in a substance, the mass of the ice cubes and the water will be equal because the same amount of matter is in the beaker.

Learn more: brainly.com/question/25150590

4 0
2 years ago
Which climate condition will cause the fastest chemical weathering of granite sandstone and shale
Naily [24]

Answer:

1. Tropical Climate

Explanation:

Due to the high heat in tropical climates, the moisture speeds up chemical weathering

6 0
3 years ago
Read 2 more answers
What is the molecular formula of the hydrocarbon whose molar mass is 536 g/mol and contains 89.55 w/w% carbon?
leonid [27]

Given the percentage composition of HC as C → 81.82 % and H → 18.18 %

So the ratio of number if atoms of C and H in its molecule can will be:

C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8

So the Empirical Formula of hydrocarbon is:

C 3 H 8

As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol

Now let Molecular formula of the HC be ( C 3 H 8 ) n

Using molar mass of C and H the molar mass of the HC from its molecular formula is:

( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1

Hence the molecular formula of HC is C 3 H 8

Does that help?

7 0
3 years ago
Write the chemical equation for the dissolution reaction of solid iron(iii) hydroxide in water. include the phases of all specie
Allushta [10]
When solid <span>iron (iii) hydroxide is dissolved into water, it ionizes or it dissociates into ions. These ions are the iron (iii) ions and the hydroxide ions. Iron(III) oxide is classified as a base when in aqueous solution since it produces hydroxide ions. It is a weak base so it does not completely dissociate into the solution. The dissociation equation would be:

Fe(OH)3 <-----> Fe3+ + OH-

To write a complete reaction, the reaction should be balanced wherein the number of atoms of each element in the reactant side and the product side should be equal. Also, the phases of the substances should be written. We do as follows:

</span>
Fe(OH)3 (s)  <-----> Fe3+ (aq) + 3OH- (aq)
4 0
3 years ago
Zinc metal and aqueous silver nitrate react to give Zn(NO3)2(aq) plus silver metal. When 5.00 g of Zn(s) and solution containing
Fittoniya [83]

Answer : The percent yield is, 83.51 %

Solution : Given,

Mass of Zn = 5.00 g

Mass of AgNO_3 = 25.00 g

Molar mass of Zn = 65.38 g/mole

Molar mass of AgNO_3 = 168.97 g/mole

Molar mass of Ag = 107.87 g/mole

First we have to calculate the moles of Zn and AgNO_3.

\text{ Moles of }Zn=\frac{\text{ Mass of }Zn}{\text{ Molar mass of }Zn}=\frac{5.00g}{65.38g/mole}=0.0765moles

\text{ Moles of }AgNO_3=\frac{\text{ Mass of }AgNO_3}{\text{ Molar mass of }AgNO_3}=\frac{25.00g}{168.97g/mole}=0.1479moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag

From the balanced reaction we conclude that

As, 2 mole of AgNO_3 react with 1 mole of Zn

So, 0.1479 moles of AgNO_3 react with \frac{0.1479}{2}=0.07395 moles of Zn

From this we conclude that, Zn is an excess reagent because the given moles are greater than the required moles and AgNO_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of Ag

From the reaction, we conclude that

As, 2 mole of AgNO_3 react to give 2 mole of Ag

So, 0.1479 moles of AgNO_3 react to give 0.1479 moles of Ag

Now we have to calculate the mass of Ag

\text{ Mass of }Ag=\text{ Moles of }Ag\times \text{ Molar mass of }Ag

\text{ Mass of }Ag=(0.1479moles)\times (107.87g/mole)=15.95g

Theoretical yield of Ag = 15.95 g

Experimental yield of Ag = 13.32 g

Now we have to calculate the percent yield.

\% \text{ yield}=\frac{\text{ Experimental yield of }Ag}{\text{ Theretical yield of }Ag}\times 100

\% \text{ yield}=\frac{13.32g}{15.95g}\times 100=83.51\%

Therefore, the percent yield is, 83.51 %

7 0
3 years ago
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