300 °C would be found in which temperature range?

PLEASE HELP ME!!!

kondor19780726 [428]

<h3>Answer:</h3>

0.424 J/g °C

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Thermochemistry</u>

Specific Heat Formula: q = mcΔT

q is heat (in Joules)
m is mass (in grams)
c is specific heat (in J/g °C)
ΔT is change in temperature

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

<u>Step 2: Solve for Specific Heat</u>

Substitute in variables [Specific Heat Formula]: 181 J = (38.8 g)c(11.0 °C)
Multiply: 181 J = (426.8 g °C)c
[Division Property of Equality] Isolate <em>c</em>: 0.424086 J/g °C = c
Rewrite: c = 0.424086 J/g °C

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.424086 J/g °C ≈ 0.424 J/g °C

6 0

2 years ago

decomposition of hydrogen peroxide produce water and oxygen . calculate the volume of O2 formed from the decomposition of 150 mL

Ugo [173]

Explanation:

2H2O2 => 2H2O + O2

Moles of hydrogen peroxide = 0.150dm³ * (0.02mol/dm³) = 0.003mol .

Moles of oxygen = 0.0015mol.

Volume of oxygen = 0.0015mol * (22.4dm³/mol) = 0.0336dm³.

5 0

3 years ago

The chemical mixture that composes our atmosphere is called ________.

Virty [35]

Answer: Air

Hope this helps

3 0

3 years ago

The conversation of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-⁴s-¹. a) i

Rus_ich [418]

Answer: a) The concentration after 8.8min is 0.17 M

b) Time taken for the concentration of cyclopropane to decrease from 0.25M to 0.15M is 687 seconds.

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

a) concentration after 8.8 min:

$8.8\times 60s=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{a-x}$

$\log\frac{0.25}{a-x}=0.15$

$\frac{0.25}{a-x}=1.41$

$(a-x)=0.17M$

b) for concentration to decrease from 0.25M to 0.15M

$t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\log\frac{0.25}{0.15}\\\\t=\frac{2.303}{6.7\times 10^{-4}s^{-1}}\times 0.20$

$t=687s$

7 0

3 years ago

Please help 16-20 thanks

MaRussiya [10]

16. FALSE
17. TRUE
18. FALSE
19. TRUE
20. TRUE

3 0

3 years ago