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3 years ago

A reaction is shown below.

Chemistry

2 answers:

Anna11 [10]3 years ago

8 0

I think it would be B "A heterogeneous catalyst is being used in the reaction"

This is because MnO2 is a heterogeneous catalyst as it's in a different state than the rest of the reactants and products.
The MnO2 isn't being used as a reactant so it's a catalyst.
MnO2 is not an enzyme

hope that helps

umka21 [38]3 years ago

7 0

Answer:

Explanation:

I took the test

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Darwin observed how farmers and breeders produced many kinds of farm animals and plants. These plants and animals had traits tha

NISA [10]

the answer is A......

it is supported by practical evidence and examples. this is the answer because he tried and tested many different ways to see what would happen so he is happy with the conclusion that what he tested is what he gets.

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4 years ago

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Rate = K[A]

Aleks [24]

Answer:

9 × 10⁻³ mol·L⁻¹s⁻¹

Explanation:

Data:

k = 1 × 10⁻³ L·mol⁻¹s⁻¹

[A] = 3 mol·L⁻¹

Calculation:

rate = k[A]² = 1 × 10⁻³ L·mol⁻¹s⁻¹ × (3 mol·L⁻¹)² = 9 × 10⁻³ mol·L⁻¹s⁻¹

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3 years ago

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

Answer: The mass of glucose in final solution is 1.085 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

1. Describe the motion of gas particles.

kykrilka [37]

Answer:

The particles in a solid are tightly packed and locked in place. ... The particles in a liquid are close together (touching) but they are able to move/slide/flow past each other. The particles in a gas are fast moving and are able to spread apart from each other.

Explanation:

3 0

3 years ago

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"What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0"

Artyom0805 [142]

Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)

Answer:

-3670.33 J/K

Explanation:

Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.

Mathematically, change of Entropy can be expressed as,

ΔS = ΔH/T ....................................... Equation 1

Where ΔS = Change of entropy, ΔH = heat change, T = temperature.

ΔH = -(Lf×m).................................... Equation 2

Note: ΔH is negative because heat is lost.

Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg

Substitute into equation

ΔH = -(3.34×10⁵×3.0)

ΔH = - 1002000 J.

But T = 0 °C = (0+273) K = 273 K.

Substitute into equation 1

ΔS = -1002000/273

ΔS = -3670.33 J/K

Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C

4 0

3 years ago