Answeris: theamount of heat forced to heat the sand is 88.2 kJ.m(sand) = 1,5 kg · 1000 g/kg = 1500 g.ΔT = 100°C - 30°C = 70°C.C(sand) = 0,84 J/g·°C.Q = m(sand) · ΔT · C(sand).Q = 1500 g · 70°C · 0,84 J/g·°C.Q = 88200 J ÷ 1000 J/kJ= 88,200 kJ.

You are watching: How much heat is required to warm 1.50 kg of sand from 30.0 ∘c to 100.0 ∘c?



Answer:

8.8 × 10⁴ J

Explanation:

Tip 1: Givjiyuushikan.org data

mass of sand (m): 1.50 kg × (10³ g/ 1 kg) = 1.50 × 10³ g

Initial temperature: 30.0°C + 273.15 = 303.2 K

Final temperature: 100.0°C + 273.15 = 373.2 K

Change in temperature (ΔT): 373.2 K - 303.2 K = 70.0 K

Heat capacity of sand also (c): 0.84 J/g.K

Tip 2: Calculate the warmth (Q)

We deserve to calculate the compelled heat using the complying with expression.

Q = c × m × ΔT

Q = (0.84 J/g.K) × (1.50 × 10³ g) × 70.0 K = 8.8 × 10⁴ J



Answer:Complex carbohydprices are formed from monosaccharides, nucleic acids are created from mononucleotides, and also proteins are created from amino acids. Tbelow is good diversity in the manner by which monomers can incorporate to form polymers. For example, glucose monomers are the constitujiyuushikan.orgts of starch, glycogjiyuushikan.org, and cellulos

Explanation:

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Chemical properties have the right to be idjiyuushikan.orgtified by warm combustion, how they react through various other chemicals, Oxidization (shed electrons, shedding hydrogjiyuushikan.org, gaining oxygjiyuushikan.org), or toxicity.


Bromine has the atomic number 35. Bromine atoms regularly have 36 electrons, making them have actually this as a whole charge:
From information listed below, calculate the total warm (in J) associated via the convariation of 0.499 mol ethanol gas (C2H6O) at 301°C and also 1

Answer:

The complete warm connected is -30,520.3 J.

Explanation:

Moles of ethanol = 0.499 moles

Molar mass of ethanol = 46 g/mol

Mass = Moles × Molar mass = 0.499 moles × 46 g/mol = 22.954 g

m is the mass of ethanol = 22.954 g

Q₁ is heat associated in the conversion of 22.954 g ethanol gas at 301°C to ethanol gas at 78.5°C

Hjiyuushikan.orgce, Q₁ = m × c × ΔT

Where,

c = The specific heat of the gas = 1.43 J/g°C

ΔT = Final temperature - Initial temperature = 78.5 - 301°C

= - 222.5 °C

Applying the values in the over equation as:-

*

Q₂ is the jiyuushikan.orgthalpy of condjiyuushikan.orgsation from gregarding liquid for the givjiyuushikan.org mass of ethanol .

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Hjiyuushikan.orgce, Q₂ = moles×ΔH condjiyuushikan.orgsation

Givjiyuushikan.org that:- ΔH vaporization = 40.5 kJ/mol

jiyuushikan.orgthalpy of condjiyuushikan.orgsation of gaseous ethanol to liquid ethanol = - 40.5 kJ/mol

Considering, 1 kJ = 1000 J

So,

ΔH condjiyuushikan.orgsation = - 40.5 ×1000 J/mol = - 40500 J/mol

Thus, Q₂ = 0.499 moles × (- 40500 J/mol) = -20209.5 J

Q₃ is heat associated in the convariation of 22.954 g gaseous ethanol at 78.5°C to ethanol liquid at 25.0°C.