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IRINA_888 [86]
3 years ago
10

In a first-order decomposition reaction. 58.6% of a compound decomposes in 11.7 min. How long (in min) does it take for 80.2% of

the compound to decompose
Chemistry
1 answer:
gavmur [86]3 years ago
6 0

Answer:

t = 21.5 min.

Explanation:

Hello!

In this case, since the kinetics of a first-order reaction is:

\frac{[A]}{[A]_0}=exp(-kt)

Thus, since we are given the 11.7 min for a 58.6-% consumption, we can compute the rate constant, k:

ln(1-0.586)=-kt\\\\k=\frac{ln(0.414)}{-t}=\frac{-0.882}{11.7min}=0.0754min^{-1}

Now, for the second problem, as the new consumption is 80.2%, we can compute the required time as shown below:

ln(1-0.802)=-kt\\\\t=\frac{ln(198)}{k} \\\\t=\frac{-1.62}{0.0754min^{-1}}\\\\t=21.5min

Best regards!

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Imagine you are about to investigate what happens when sodium and potassium are mixed together. Before you begin your experiment
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Explanation:

Potassium and Sodium are both metals on the periodic table. Both elements are found on the s-block of the table. Therefore, it is not possible for both of them to combine to form a compound.

For a compound to form two elements must combine in a definite order to given  single product.

The combination is facilitated by a loss, gain or sharing of electrons between two species. This leads to an attraction between the combining species.

Both Sodium and potassium would prefer to lose electrons and there is no reason for them to combine to form a compound.

6 0
3 years ago
Suppose you begin with an unknown volume of 8.61 m h2so4 and add enough water to make 5.00*102 ml of a 1.75 m h2so4 solution. wh
olga55 [171]
<span>1.02x10^2 ml Since molarity is defined as moles per liter, the product of the molarity and volume will remain constant as mole solvent is added. So let's set up an equality to express this m0*v0 = m1*v1 where m0, v0 = molarity and volume of original solution m1, m1 = molarity and volume of final solution. Solve for v0, then substitute the known values and calculate: m0*v0 = m1*v1 v0 = (1.75 M * 500 ml)/8.61 M v0 = (1.75 M * 500 ml)/8.61 M V0 = 101.6260163 Rounding to 3 significant figures gives 102 ml. So the original volume of the 8.61 M H2SO4 solution was 102 ml or 1.02x10^2 ml.</span>
3 0
3 years ago
Prior to hosting an international soccer match, the local soccer club needs to replace the artifical turf on their field with gr
GuDViN [60]

Answer:

It will cost $68,620.5 to the club to add the grass turf to their field.

Explanation:

length of the field = l = 0.102 km = 0.102 × 1000 m

( 1km = 1000 m)

Width of the field w = 0.069 km = 0.069 × 1000 m

Area of the field , A= l × w

A=0.102\times 1000 m\times 0.069\times 1000 m=7,038 m^2

Cost of grass turfing = 9.75 \$/m^2

Cost of grass turfing on field of 7,038 m^2 :

=7,038 m^2\times 9.75 \$/m^2=\$68,620.5

It will cost $68,620.5 to the club to add the grass turf to their field.

4 0
3 years ago
I need help on this someone please help
Murljashka [212]

1) Dawn dish soap has a density of 1.06 g/mL. If the mass of a sample of the liquid is 1.00 g what is the volume?

Answer:

v = 0.94 mL

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Given data:

Density of soap =  1.06 g/mL.

Mass  = 1 g

Volume = ?

Solution:

d = m/v

v = m/d

v = 1 g/1.06 g/mL

v = 0.94 mL

2) Maple syrup has a density of 1.37 g.mL. What is the mass of 1.0 L of the maple syrup?

Answer:

m = 1370 g

Given data:

Density of soap =  1.37 g/mL.

Mass  = ?

Volume = 1.0 L ( 1000 mL)

Formula:

D=m/v

D= density

m=mass

V=volume

Solution:

d = m/v

m = d × v

m = 1.37 g/mL  ×  1000 mL

m = 1370 g

3) The density of gasoline is 0.754 g/mL. A drop of gasoline has a mass of 22 g what is the volume?

Answer:

v = 29.2 mL

Given data:

Density of soap =  0.754 g/mL.

Mass  = 22 g

Volume = ?

Formula:

D=m/v

D= density

m=mass

V=volume

Solution:

d = m/v

v = m/d

v = 22 g/0.754 g/mL

v = 29.2 mL

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defon

Answer:

Any kind, as long as there is an action.

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