1.Light waves (shown) are an example of what type of wave? (Lesson 4.03)

I need help what is s-1/3=7/9???? please help guys and thank you

nalin [4]

Answer:

10/9

Explanation:

First, let's convert 1/3 and 7/9 so that the have the same denominator. To do this let's find the least common multiple of 3 and 9.

List the multiples of 3 and 9:

3: 3, 9

9: 9

They have a least common multiple of 9

We need to convert 1/3 so it has a denominator of 9:

1/3*3/3 (we can multiply it by 3/3 because any number over itself is 1) = 3/9

s-3/9=7/9

Add 3/9 to both sides to isolate s

s=10/9

6 0

2 years ago

A sample of hydrogen was collected by water displacement at 23.0°C and an atmospheric pressure of 735 mmHg. Its volume is 568 mL

ziro4ka [17]

Answer:

V = 552 mL or 0.552 L

Explanation:

First, we need to calculate the number of moles of H2 using the ideal gas equation which is:

PV = nRT

Solving for n:

n = PV / RT

Where:

P = Pressure

V = Volume

R = Gas constant (0.082 L atm / K mol)

T = Temperature in K

Let's convert first both pressure in atm, remember that 1 atm = 760 mmHg

P = 735 / 760 = 0.967 atm

Pwater = 21 / 760 = 0.028 atm

Finally temperature to Kelvin:

T = 23 + 273.15 = 296.15 K

Now, at first the hydrogen was collected by water displacement so pressure is:

P = 0.967 - 0.028 = 0.939 atm

Now the moles of hydrogen:

n = 0.939 * 0.568 / 0.082 * 296.15

n = 0.022 moles

Now that we have the moles, let's calculate the volume when the pressure is 735 mmHg

V = nRT/P

V = 0.022 * 0.082 * 296.15 / 0.967

V = 0.552 L or 552 mL

This is the volume that hydrogen occupies.

6 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

What law represents a balanced chemical equation

PSYCHO15rus [73]

Answer:

The law of conservation of mass represents a balanced chemical equation.

8 0

2 years ago

Read 2 more answers

If 0.600mol of chloride gas reacted with 0.500mol of aluminium metal to produce aluminium chloride,which reactant is in excess?h

Rus_ich [418]

Answer:

The balanced chemical equation: 2 Al + 3Cl2→ 2 AlCl3

Mole-mole relationship: 2 moles Al + 3 moles Cl2→ 2 moles AlCl3

Given: 0.600 moleCl2; 0.500 mole Al

Required: Excess reactant___; Number of moles of AlCl3 produced__

Solution: Use dimensional analysis using the mole-mole rel

0.600 mole Cl2 * 2 moles Al/3 moles Cl2 = 0.4 mole Al

0.5 mole Al* 3 moles Cl2/2 moles Al = 0.75 mole Cl2

Based on the given:

0.6mole Cl2 + 0.4 mole Al ( this is possible based on the given)

0.5mole Al + 0.75 mole Cl2 (this is not possible because the given is only 0.600 mole of Cl 2)

Answer: Excess reactant is Al; Limiting reactant is Cl2

The amount of AlCl3 produced = 0.6 mole Cl2 + 0.4 mole Al = 1.0 mole AlCl3

4 0

2 years ago