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Natasha_Volkova [10]
3 years ago
12

In the chemical reaction:

Chemistry
1 answer:
Fiesta28 [93]3 years ago
8 0
The correct answer is A. AgCl (s) only
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B. Determine the percent composition of each element in NaCl. (1 point) Must show work to
andrey2020 [161]

Answer:

Sodium (Na): (.5 point)

22.99+35.453=58.433.   22.99/58.433=   39%

Chlorine (Cl): (.5 point)

22.99+35.453=58.433.   35.453/58.433= 61%

Explanation:

8 0
3 years ago
Help now im not trying to fail
Alekssandra [29.7K]

Answer:

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5 0
3 years ago
Read 2 more answers
how much hydrogen will be released during the interaction of potassium weighing 8 grams with hydrochloric acid
krek1111 [17]

Answer:

With an understanding of the ideal gas laws, it is now possible to apply these principles to chemical stoichiometry problems. For example, zinc metal and hydrochloric acid (hydrogen chloride dissolved in water) react to form zinc (II) chloride and hydrogen gas according to the equation shown below:

2 HCl (aq) + Zn (s) → ZnCl2 (aq) + H2 (g)

Explanation:

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4 0
3 years ago
How many molecules of CBr4 are in 250 grams of CBr4
Kazeer [188]

Answer:- 4.54*10^2^3 molecules.

Solution:- The grams of tetrabromomethane are given and it asks to calculate the number of molecules.

It is a two step unit conversion problem. In the first step, grams are converted to moles on dividing the grams by molar mass.

In second step, the moles are converted to molecules on multiplying by Avogadro number.

Molar mass of CBr_4  = 12+4(79.9)  = 331.6 g per mol

let's make the set up using dimensional analysis:

250g(\frac{1mol}{331.6g})(\frac{6.022*10^2^3molecules}{1mol})

= 4.54*10^2^3 molecules

So, there will be 4.54*10^2^3 molecules in 250 grams of CBr_4 .


8 0
3 years ago
Transuranium elements are artificial and may be prepared by a fusion process in which heavy nuclei are bombarded with light ones
Scilla [17]

Answer:

_{97}^{345}\textrm{Bk}

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

Am has an atomic number of 95. So correct reaction is:-

^{343}_{95}\textrm{Am}+^{4}_{2}\textrm{He}\rightarrow ^A_Z\textrm{X}+2^1_0\textrm{n}

To calculate A:

Total mass on reactant side = total mass on product side

343 + 4 = A + 2

A = 345

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

95 + 2 = Z + 0

Z = 97

Hence, the isotopic symbol of unknown element is _{97}^{345}\textrm{Bk}

7 0
3 years ago
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