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Elenna [48]
3 years ago
14

Write a balanced equation for the double-replacement precipitation reaction described, using the smallest possible integer coeff

icients. A precipitate forms when aqueous solutions of ammonium bromide and silver(I) nitrate are combined. (Use the lowest possible coefficients. Omit states of matter.)
Chemistry
1 answer:
ahrayia [7]3 years ago
6 0

Answer:

NH4Br + AgNO3 —> AgBr + NH4NO3

Explanation:

When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:

NH4Br(aq) + AgNO3(aq) —>

In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:

NH4Br(aq) —> NH4+(aq) + Br-(aq)

AgNO3(aq) —> Ag+(aq) + NO3-(aq)

The double displacement reaction will occur as follow:

NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)

NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)

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What is the limiting reagent when 49.84 g of nitrogen react with 10.7 g of hydrogen to make ammonia
IgorLugansk [536]
If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.

We solve this problem by first writing the equation
N2 + 3H2 = 2NH3
N2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g
so 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent
8 0
3 years ago
Determine the mass of carbon iv oxide ,produced on burning 104g of ethyne​
mars1129 [50]

163 grams (3 sig. fig.).

<h3>Explanation</h3>
  • Formula of <em>carbon(IV) oxide</em> (a.k.a. carbon dioxide): \text{CO}_2.
  • Molar mass of \text{CO}_2: \underbrace{12.01}_{\text{C}} + 2\times\underbrace{16.00}_{\text{O}}=44.01\;\text{g}\cdot\text{mol}^{-1}.
  • Formula of ethyne: structural \text{H}-\text{C}\equiv\text{C}-\text{H} or molecular \text{C}_2\text{H}_2.
  • Molar mass of \text{C}_2\text{H}_2: 2\times\underbrace{12.01}_{\text{C}}+2 \times\underbrace{16.00}_{\text{O}} = 56.02\;\text{g}\cdot\text{mol}^{-1}.

All carbon atoms in that 104 grams of ethyne will end up in \text{CO}_2. Number of moles of molecules in 104 grams of ethyne:

n = \dfrac{m}{M} = \dfrac{104}{56.02} = 1.85648\;\text{mol}.

There are two carbon atoms in each ethyne molecule. Number of carbon atoms in that many ethyne molecules:

n(\text{C}) = 2\;n(\text{C}_2\text{H}_2) = 3.71296\;\text{mol}.

There are one carbon atom in each \text{CO}_2 molecule. In case oxygen is in excess, all those carbon atoms from that 104 grams of ethyne will make n(\text{CO}_2) = n(\text{C}) =3.71296\;\text{mol} of \text{CO}_2.

Mass of all those \text{CO}_2 molecules:

m = n\cdot M = 163\;\text{g}. (3 sig. fig. as in the mass of ethyne.)

7 0
3 years ago
Please help-<br> thank you-
Flauer [41]
Dang that’s crazy.. Goodluck ..
4 0
3 years ago
Read 2 more answers
PLEASE HELP DUE SOON
jonny [76]

Answer:

This is all true if the atom has to be neutral.

Also what does V mean?

Helium: one shell with 2 neutrons and 2 protons in the center, with 2 electrons in the first shell.

Lithium: two shells with 4 neutrons and 3 protons in the center, with 2 electrons in the first shell, and 1 electron in the second shell.

Nitrogen: two shells with 7 neutrons and 7 protons in the center, with 2 electrons in the first shell, and 5 electrons in the second shell.

Flourine: two shells with 9 protons and 10 neutrons in the center, with 2 electrons in the first shell, and 7 electrons in the second shell.

Neon: two shells with 10 neutrons and 10 protons in the center, with 2 electrons in the first shell, and 8 electrons in the second shell.

Boron: two shells with 6 neutrons and 5 protons in the center, with 2 electrons in the first shell, and 3 electrons in the second shell.

3 0
3 years ago
Calculate the mass (in grams) of 8.56 moles of sulfur.
oksano4ka [1.4K]

Answer:

275g

Explanation:

Depending on the molar mass you are given, you can use that to solve this.

(I'm going based on my science class' molar mass of sulphur being 32.07g/mol)

Starting off, the formula for finding moles is

n=m/M (moles = mass / molar mass)

We can manipulate this equation to solve for mass.

m=Mn

now fill in what we now.

m = 32.07*8.56

mass = 274.5192

Now round for significant digits (if you are needed to do)

mass = 275g

5 0
3 years ago
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