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nalin [4]
4 years ago
11

What is the ionic equation of Aqueous calcium bromide was mixed with aqueous gold(I) perchlorate, and a crystallized gold(I) bro

mide product was formed.?
Chemistry
1 answer:
Bumek [7]4 years ago
6 0

Answer:- 2AuClO_4(aq) + CaBr_2(aq)\rightarrow 2AuBr(s)+ Ca(ClO_4)_2(aq)

Explanations:- It's a double replacement reaction where a precipitate of silver(I)bromide is formed. A dpuble replacement reaction in general looks as AB + CD \rightarrow AD + CB

To balance the equation we need to multiply gold compounds on both sides by 2 and the balanced equation is..

2AuClO_4(aq) + CaBr_2(aq)\rightarrow 2AuBr(s)+ Ca(ClO_4)_2(aq)

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Which of the following would increase the strength of an electromagnet? Check all that apply.
TiliK225 [7]

Answer: The correct statements are ; A and B.

Explanation:

The strength of the magnetic field(B) in an electromagnet can be calculated using formula:

B=\mu\frac{N}{L}I

\mu = Relative permeability

N = number of turns

I = Current in the wire of the solenoid

L = length of the solenoid or electromagnetic

As we can see from the formula:

B\propto N

B\propto I

So, by increasing the turns and increasing current flowing through wire one can increase the strength of an electromagnet.

Hence, the correct statements are ; A and B.

5 0
3 years ago
State two essential conditions for rusting of iron. List two methods of protecting iron from rusting?​
gizmo_the_mogwai [7]
Conditions for rusting
• moisture
•oxygen

Prevention
•Painting
• Galvanisation
• Making Alloys
• greasing
Hope this helps you ;)(:

8 0
3 years ago
Elements are simple ___
storchak [24]

Answer:

particles

atoms

broken

isotope

conductors

electricity

element

combined

3 0
3 years ago
If 61.5 moles of an ideal gas occupies 97.5 liters at 473 K, what is the pressure of the gas, in mmHg?
kiruha [24]
I’d say for the answer 13.13 mmHg?
5 0
3 years ago
It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis
Norma-Jean [14]

Answer:

a)

The overall  balanced combustion  reaction is written as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

(F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = 23.562

b)

the higher heating values (HHV)_f per unit mass of LPG = 49.9876 MJ/kg

the lower heating values (LHV)_f per unit mass of LPG = 46.4933 MJ/kg

Explanation:

a)

The stoichiometric equation can be expressed as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> aCO_2  \ + \ bH_2O \ + \ cN_2

Now, equating the coefficient of carbon; we have:

(0.7×3)+(0.05×4)+(0.25×3) = a

a = 3.05

Also, Equating the coefficient of hydrogen : we have:

(0.7 × 8) +(0.05 × 10) + ( 0.25 × 6) = 2 b

2b = 7.6

b = 3.8

Equating the coefficient of oxygen

2x = 2a + b

x = \frac{2a+b}{2} \\ \\ x =  \frac{2(3.05)+3.8}{2} \\ \\ x = 4.95

Equating the coefficient of Nitrogen

c = 3.76x \\ \\ c = 3.76 *4.95 \\ \\ c = 18.612

Therefore, The overall  balanced combustion  reaction can now be written as :

0.7C_3H_8 \ + \ 0.05C_4H_{10} \ + \ 0.25 C_3H_6 \ + \ x(O_2 \ + \ 3.76N_2) ---> 3.05CO_2  \ + \ 3.8H_2O \ + \ 18.612N_2

Now;  To determine the stoichiometric F/A and A/F ratios; we have:

(F/A)_{stoichiometric} = \frac{n_f}{n_a } \\ \\  (F/A)_{stoichiometric} = \frac{1}{4.95*(1+3.76)} \\ \\ (F/A)_{stoichiometric} = 0.0424

(A/F)_{stoichiometric} = \frac{n_a}{n_f } \\ \\  (A/F)_{stoichiometric} = \frac{4.95*(1+3.76)}{1} \\ \\ (A/F)_{stoichiometric} = 23.562

b)

What are the higher and lower heating values per unit mass of LPG?

Let calculate the molecular mass of the fuel in order to determine their mass fraction of the fuel components.

Molecular mass of the fuel M_f = (0.7*M_{C_3H_5} ) + (0.05 *M_{C_4H_{10}}) + (0.25*M _{C_3H_6})

= 30.8 + 2.9 + 10.5

= 44.2 kg/mol

Mass fraction of the fuel components can now be calculated as :

m_{C_3H_8} = \frac{30.8}{44.2} \\ \\ m_{C_3H_8}  = 0.7 \\ \\ \\  m_{C_4H_{10}} = \frac[2.9}{44.2} \\ \\ m_{C_4H_{10}} = 0.06  \\  \\ \\ m_{C_3H_6} = \frac{10.5}{44.2} \\ \\ m_{C_3H_6}  = 0.24

Finally; calculating the higher heating values (HHV)_f per unit mass of LPG; we have:

(HHV)_f=(0.7 * HHV_{C_3H_8}) + (0.06 *HHV_{C_4H_{10}})+(0.24*HHV_{C_3H_6} \\ \\ (HHV)_f=(0.7*50.38)+(0.06*49.56)+(0.24*48.95) \\ \\ (HHV)_f=49.9876 \ MJ/kg

calculating the lower heating values (LHV)_f per unit mass of LPG; we have:

(LHV)_f = (HHV)_f - \delta H_w \\ \\  (LHV)_f = (HHV)_f  - [\frac{m_w}{m_f}h_{vap}] \\ \\ (LHV)_f   = 49.9876 \ MJ/kg -  [\frac{3.8*18}{44.2}*2.258 \ MJ/kg]  \\ \\ (LHV)_f = 46.4933 \ M/kg

7 0
4 years ago
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