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Brut [27]
3 years ago
12

Where does washing powder fall into in the ph scale ?

Chemistry
1 answer:
JulijaS [17]3 years ago
8 0
The powder is made through a chemical reaction (saponification) between an alkaline base with oils is to be expected that there will always be a residual alkali. PH around <span>11.5

hope this helps!

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Consider the following thermochemical reaction for kerosene:
ch4aika [34]

We have to solve this question using the stoichiometry of the reaction:

The equation of the reaction is;

2 C12H26(l) + 37 O2(g) -----> 24 CO2(g) + 26 H2O(l) + 15,026 kJ

According to the question;

Number of moles of CO2 released = 21.3 g/44 g/mol = 0.48 moles

From the  stoichiometry of the reaction:

Since;

24 moles of CO2 released 15,026 KJ

0.48 moles of CO2 will release 0.48 * 15,026/24

= 301 KJ of heat.

brainly.com/question/6901180

7 0
3 years ago
Read 2 more answers
The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other plati
Alinara [238K]

Here we have to identify the metal powder by the given disposal.

The identification of Zinc can be done by 1 m nitric acid (HNO₃) and Ni(NO₃)₂ which will produce hydrogen gas by reaction and displacement reaction as shown below.

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

The identification of lead can be done by the reaction with 1 m nitric acid (HNO₃) which produces lead nitrate.

The reaction is-

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

The identification of platinum can be done by the reaction with all the given disposal as it will not react with any of the compound.

1. Identification of Zinc (Zn):

(a) Zn metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than zinc.

Zn + NaNO₃ = No reaction

(b) Zn metal will react with 1 m HNO₃ to form hydrogen gas. The reaction is:

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

(c) Zn will react with nickel nitrate [Ni(NO₃)₂] because it may only cause displacement reaction the reduction potential of Zn²⁺/Zn (-0.76) is less than that of Ni²⁺/Ni (-0.23). Thus the reaction will be:

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

2. Identification of lead (Pb):

(a) Pb metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pb.

Pb + NaNO₃ = No reaction

(b) Pb reacts with HNO₃ to form lead nitrate. The reaction is:

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

(c) The standard reduction potential of Pb²⁺/Pb is more than nickel Ni²⁺/Ni thus there will be no reaction between Pb and NI(NO₃)₂.

Pb + Ni(NO₃)₂ = No reaction.

3. Identification of platinum (Pt)

(a) Pt metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pt.

Pt + NaNO₃ = No reaction.

(b) The standard reduction potential of Pt²⁺/Pt is so high (+1.188) thus there will be no reaction with HNO₃.

Pt + HNO₃ = No reaction

(c) The standard reduction potential of Pt²⁺/Pt is more than nickel Ni²⁺/Ni thus there will be no reaction between Pt and Ni(NO₃)₂.

Pt +  Ni(NO₃)₂ = No reaction.    

8 0
3 years ago
I need yall help with this one and i can make yall brilientist if u get it right plus yall get 50 points
mixas84 [53]

Answer:

D

Explanation:

6 0
2 years ago
Read 2 more answers
Please solve this fast pleasee i will make u the brainliest
mote1985 [20]
1. Thermoplastics
2. Thermosetting plastics
4. Nylon
5. Thermoplastics like polythene
6. Thermosetting plastics like melamine
7. Natural
8. Raw
11. Light and durable
13. repeating subunits
14. Naturally from silkworms
15. Thermoplastics
16. Air and water
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3 0
3 years ago
60 grams of ice will require _____ calories to raise the temperature 1°C.
guajiro [1.7K]

Answer:

60 grams of ice will require 30.26 calories to raise the temperature 1°C.

Explanation:

The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released or absorbed by the system.

m is the mass of the ice (m = 60.0 g).

c is the specific heat capacity of ice (c = 2.108 J/g.°C).

ΔT is the temperature difference (ΔT = 1.0 °C).

∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.

<em>It is known that 1.0 cal = 4.18 J.</em>

<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>

7 0
3 years ago
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