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Fudgin [204]
2 years ago
8

What is the result of adding 2.5 × 103 and 3.5 × 102?

Chemistry
2 answers:
dangina [55]2 years ago
6 0
2.5×103 and 3.5 x 102=6.0x103
aleksklad [387]2 years ago
4 0

Answer:

The correct answer is option (a).

Explanation:

(2.5\times 10^3)+(3.5\times 10^2

Taking 10^2 common form both brackets:

=10^2((2.5\times 10)+(3.5\times 1))

Taking 5 common form both brackets:

=10^2\times 5((0.5\times 10)+(0.7\times 1))

=10^2\times 5(5.0+0.7)=10^2\times (28.5)

=2.85\times 10^3\approx 2.9 10^3[

Hence,the correct answer is option (a).

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When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be
Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH  = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL  +  50.mL  = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH_{neutralization}

ΔH_{neutralization} = -q_soln / mole of water produced

so we substitute

ΔH_{neutralization} = -( 4076.68 J ) / 0.0500 mol  

ΔH_{neutralization} = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

6 0
2 years ago
The lithosphere, hydrosphere, and ______________ are the abiotic parts of the planet.
VladimirAG [237]
The lithosphere, hydrosphere, and atmosphere are the abiotic parts of the planet.
8 0
3 years ago
What is the volume of kristas rock
Elan Coil [88]

Answer:

for what I can see in the picture the volume is 155

Explanation:

7 0
2 years ago
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If 25.5g of sodium thiosulphate was dissolved in 40g of distilled water at 25°C,
zhenek [66]

Answer:

Formula: Na2S2O3

we get solubility.

Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .

Solution given:

mass of sodium thiosulphate [m1]=25.5g

mass of water [m2]=40g

at temperature [t]=25°C

we have

<u>solubility in g/dm^3</u> :\frac{solute in gram}{solvent in gram} *100

  • =\frac{25.5}{40}*100
  • =63.75g /litre=63.75g/dm³

<u>solubility in g/dm^3 :63.75g/dm³</u>

<u>n</u><u>o</u><u>w</u>

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Which is NOT a chemical reaction?
Sophie [7]
A is correct because there is no reaction involved
6 0
3 years ago
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