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3 years ago

Compounds have physical properties that

1 answer:

4 0

The answer is B, they differ greatly from the properties of all their elements. For example: Water is non-flamable but it is made from hydrogen, which is highly flammable, and oxygen, a gas that supports combustion.

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A food substance kept at 0Â°C becomes rotten (as determined by a good quantitative test) in 8.3 days. The same food rots in 10.6

ZanzabumX [31]

Answer:

1. 67.2 kJ/mol

Explanation:

Using the derived expression from Arrhenius Equation

$In \ (\frac{k_2}{k_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})$

Given that:

time $t_1$ = 8.3 days = (8.3 × 24 ) hours = 199.2 hours

time $t_2$ = 10.6 hours

Temperature $T_1$ = 0° C = (0+273 )K = 273 K

Temperature $T_2$ = 30° C = (30+ 273) = 303 K

Rate = 8.314 J / mol

Since $(\frac{k_2}{k_1}=\frac{t_2}{t_1})$

Then we can rewrite the above expression as:

$In \ (\frac{t_2}{t_1}) = \frac{E_a}{R}(\frac{T_2-T_1}{T_2*T_1})$

$In \ (\frac{199.2}{10.6}) = \frac{E_a}{8.314}(\frac{303-273}{273*303})$

$2.934 = \frac{E_a}{8.314}(\frac{30}{82719})$

$2.934 = \frac{30E_a}{687725.766}$

$30E_a = 2.934 *687725.766$

$E_a = \frac{2.934 *687725.766}{30}$

$E_a =67255.58 \ J/mol$

$E_a =67.2 \ kJ/mol$

7 0

3 years ago

A gas has a volume of 450.0 ml at standard pressure. If the temperature is held constant, what volume would the gas occupy if th

zimovet [89]

A. 1800L !:$:!:):&/&/?)3&2!3?372&2!3)3$3!3!3?3:$heguejeneve

6 0

3 years ago

If the volume of both A and C is 2mL, assuming the particles are the same size, what is the density of A? Remember D=M/V

Marat540 [252]

How to Inbox anyone here..

i am a new user.

7 0

3 years ago

Read 2 more answers

I need some help with a long chemistry problem. Anything is appreciated!

Mkey [24]

Answer:

The CSI is wrong.

Explanation:

1. Find the volume of the pool

The formula for the volume of a cylinder is V = πr²h .

D = 12 m; h = 10 m

r = D/2 = (12 m/2) = 6.0 m

V = πr²h = π × (6.0 m)² × 10 m = π × 36 m²× 10 m = 360π m³ = 1100 m³

= 1.1. × 10⁶ L

2. Calculate the moles of OH⁻

n = cV = 1.0 × 10⁻² mol·L⁻¹ × 1.3 × 10⁶ L = 11 000 mol of OH⁻

3. Calculate the moles of acetic acid needed for neutralization

HA + OH⁻ ⟶ A⁻ + H₂O

The molar ratio of is 1 mol HA:1 mol OH⁻, so you need 11 000 moL of acetic acid.

4. Calculate the actual moles of acetic acid

You have four 5 L jugs of acetic acid pH 2 .

Volume = 20 L

[H⁺] = 10⁻² mol·L⁻¹ = 0.01 mol·L⁻¹

(a) Set up an ICE table

HA + H₂O ⇌ A⁻ + H₃O⁺

I/mol·L⁻¹: c 0 0

I/mol·L⁻¹: - 0.01 +0.01 +0.01

I/mol·L⁻¹: c - 0.01 0.01 0.01

$K_{\text{a}} = \dfrac{\text{[A]}^{-}\text{[H$_{3}$O$^{+}$]}}{\text{[HA]}} = 1.76 \times 10^{-5}$

(b) Calculate the concentration of acetic acid

$\begin{array}{rcl}\dfrac{\text{[A]}^{-}\text{[H$_{3}$O$^{+}$]}}{\text{[HA]}}& = & 1.76 \times 10^{-5}\\\\\dfrac{0.01\times 0.01}{c}& = & 1.76 \times 10^{-5}\\\\1 \times 10^{-4} & = & 1.76 \times 10^{-5}c\\c & = & \dfrac{1 \times 10^{-4}}{1.76 \times 10^{-5}}\\\\ & = & \text{6 mol/L}\\\end{array}$

The concentration of the acetic acid is 6 mol·L⁻¹

(c) Calculate the moles of acetic acid

$n = \text{20 L} \times \dfrac{\text{6 mol}}{\text{1 L}} = \textbf{100 mol}$

You have 100 mol of acetic acid.

The CSI is wrong.

You don't have enough acetic acid to neutralize the pool.

8 0

4 years ago

Fill in the coefficients that will balance the following reaction: a0Na + a1H2O → a2NaOH + a3H2

Alexandra [31]

Hey there ! :

2 Na + 2 H2O = 2 NaOH + 1 H2

Sodium + Dihydrogen Monoxide = Natriumhydroxid + Hydrogen

Coefficients:

Reagents : Na = 2
H2O = 2

Products : NaOH = 2
H2 = 1

4 0

4 years ago

Read 2 more answers