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stepan [7]
3 years ago
12

In the reaction of Epsom Salt and ammonia, how many atoms of Hydrogen are in the product, ammonium sulfate (NH4)2SO4

Chemistry
2 answers:
pashok25 [27]3 years ago
5 0

<u>Answer:</u>

Epsom salt is Magnesium sulphate (MgSO_4)

So the chemical reaction is  

MgSO_4+2 NH_3+2H_2 O > 1(NH_4 )_2 SO_4+Mg(OH)_2  

1(NH_4 )_2 SO_4  contains 1\times 4\times 2=8 mole of H atoms

That is to find the number of atoms we multiply coefficient subscript number outside the Bracket

1 mole of H atoms contains 6.022 \times 10^23 H atoms

So,  

8 \ mole \ H \ atoms \times \frac {(6.022 \times 10^23 \ H \ atoms)}{(1 \ mole \ H \ atoms)}  =   48.176 \times 10^23  \ H \ atoms

Moving the decimal point to the right, the power of 10 decreases

So, we get  

=  4.8176 \times 10^22  \ H  \ atoms

= 4.82 \times 10^22 \ H atoms is the Answer.

Marrrta [24]3 years ago
4 0

Answer:

Should be 1

Explanation:

You might be interested in
Does anyone have any idea what this means lol
miv72 [106K]

Answer:

29.42 Litres

Explanation:

The general/ideal gas equation is used to solve this question as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K

According to the information provided in this question;

mass of nitrogen gas (N2) = 25g

Pressure = 0.785 atm

Temperature = 315K

Volume = ?

To calculate the number of moles (n) of N2, we use:

mole = mass/molar mass

Molar mass of N2 = 14(2) = 28g/mol

mole = 25/28

mole = 0.893mol

Using PV = nRT

V = nRT/P

V = (0.893 × 0.0821 × 315) ÷ 0.785

V = 23.09 ÷ 0.785

V = 29.42 Litres

6 0
3 years ago
The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions
mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0
3 years ago
A calorimeter contains 500 g of water at 25°C. You place a hand warmer containing 100 g of liquid sodium acetate (NaAC) inside t
antoniya [11.8K]
The heat absorbed by the water is
Q = 500 (4.18) (32.2 - 25)
Q = 15048 J

The enthalpy of fusion of the sodium acetate is:
<span>ΔHf = Q / m
</span><span>ΔHf = 15048 / 100
</span>ΔHf = 150.48 J/g
3 0
3 years ago
Read 2 more answers
Please and thank you!
Valentin [98]

Answer:

7.5 moles of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Finally, we shall determine the number of mole of O₂ produced by the decomposition of 5 moles of KClO₃. This can be obtained as follow:

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, 5 moles of KClO₃ will decompose to produce = (5 × 3)/ 2 = 7.5 moles of O₂.

Thus, 7.5 moles of O₂ were obtained from the reaction.

3 0
3 years ago
What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?
xz_007 [3.2K]

Answer:

B) 0.32 %

Explanation:

Given that:

K_{a}=1.8\times 10^{-5}

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}

The expression for dissociation constant of acid is:

K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}

1.8\times 10^{-5}=\frac{x^2}{1.8-x}

1.8\left(1.8-x\right)=100000x^2

Solving for x, we get:

<u>x = 0.00568  M</u>

Percentage ionization = \frac{0.00568}{1.8}\times 100=0.32 \%

<u>Option B is correct.</u>

8 0
3 years ago
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