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3 years ago

How much heat is required to warm 1.30 l of water from 24.0 ∘c to 100.0∘c? (assume a density of 1.0g/ml for the water.)?

1 answer:

5 0

When q = M* C *ΔT

we can get the mass using the given volume & density:

M = volume * density

= 1.3 L * 1g/ml * 1000mL1L

= 1300 g
and when the specific heat of water = 4.18J/g.°C

now, by substitution:

∴ q = 1300g * 4.18J/g.°C * (100-24)

= 412984 J or 412.984 KJ

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a dog is trained to sit and shake hands. These traits are most likely? a. inherited b. not inherited c. not acquired d. innate

Ksenya-84 [330]

<span>A dog is trained to sit and shake hands. These traits are most likely?

Answer: B not </span>acquired

hope this helps

4 0

3 years ago

The periodic table of the elements lists the elements in order of increasing atomic number. Each element has its own unique squa

Vesna [10]

Answer:

A) They all have 12 protons.

Explanation:

Magnesium has an atomic number of 12 which means it has 12 protons. All Magnesium atoms have 12 protons, the neutrons however may differ which produces things called isotopes where the atoms have same protons but neutrons change.

Keep in mind the atomic number is unique to each element, so 12 atomic number will always be Magnesium, 1 will always be Hydrogen and so on.....

6 0

3 years ago

2. Given the unbalanced equation:

zheka24 [161]

2 Na + 2 H2O → 2 NaOH + H2 (balanced equation)
The answer would be 2, since 2 in the coefficient of both Na and NaOH

7 0

3 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}