All categories

3 years ago

PLEASE HELP!

Chemistry

2 answers:

Firlakuza [10]3 years ago

7 0

Answer:

The number of particles able to undergo a chemical reaction is less than the number that is not able to.

Explanation:

took test

Flauer [41]3 years ago

4 0

Answer:

option 2 and 4 are most likely to happen but I could be wrong

You might be interested in

How does the density of a hershey bar compare to the density of the a half bar?

anastassius [24]

The density would be the same for the whole bar as well as one half of the bar. Density is a identity I believe, by this I mean that it stays the same no matter how little or how much of the same substance you have. Since density = mass / volume, half the bar has half of the weight as well as half of the volume of the whole bar, making the density the same.

For example, a block weighs 10 grams and has a volume of 5 ml. the density would be d = 10/5 or, d = 2g/ml

Half of the block weighs 5 grams and has a volume of 2.5 ml. The density is d = 5/2.5, or, d = 2 g/ml.

See, although there are different amounts of the same substance, their density is the same.

8 0

2 years ago

How did Mendeleev set up his Periodic Table of Elements?

Ksju [112]

He set up his periodic table by the atomic mass

8 0

2 years ago

A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.

sleet_krkn [62]

Explanation:

The given data is as follows.

Pressure (P) = 760 torr = 1 atm

Volume (V) = $720 cm^{3}$ = 0.720 L

Temperature (T) = $25^{o}C$ = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

PV = nRT

Total atoms present (n) = $\frac{PV}{RT}$

= $1 \times \frac{0.720 L}{0.0821 \times 298}$

= 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

x + y = 0.0294

Also, 40x + 131y = 2.966

x = 0.0097 mol

y = (0.0294 - 0.0097)

= 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = $\frac{y}{(x+y)}$

= $\frac{0.0197}{0.0294}$

= 0.67

Therefore, the mole fraction of Xe is 0.67.

6 0

2 years ago

What is 2KNO3 + 1H2CO3 = 1K2CO3+2HNO3 reaction type

Tema [17]

DR is the correct answer

hope this helps

4 0

2 years ago

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r

rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

= 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

$K=Ae^{\frac{-E}{RT}}$

If $K_1=Ae^{\frac{-E_1}{RT}} -------equation 1$ &

$K_2=Ae^{\frac{-E_2}{RT}} -------equation 2$

$\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}$

$E_1= E_2-RT*In(\frac{K_1}{K_2})$

$E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})$

$E_1 = 28957.39292$ $J/mol$

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0

2 years ago