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2 years ago

I need help, someone please help me

Chemistry

2 answers:

MArishka [77]2 years ago

6 0

Answer:

c) the panda eats both plant and animal: fish and bamboo

d)it shows direction of feeding, the one below is fed on by the one above

zimovet [89]2 years ago

3 0

C) eats both plants and animals/meats
D) how everything is related

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the concentration of the radio active isotope potassium-40 in a rock sample is found to be 6.25%. what is the age of the rock

julsineya [31]

Answer:

5.0 x 10⁹ years.

Explanation:

It is known that the decay of a radioactive isotope isotope obeys first order kinetics.

Half-life time is the time needed for the reactants to be in its half concentration.

If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).

Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

The half-life of K-40 = 1.251 × 10⁹ years.

For, first order reactions:

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1.251 × 10⁹ years) = 5.54 x 10⁻¹⁰ year⁻¹.

Also, we have the integral law of first order reaction:

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 5.54 x 10⁻¹⁰ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of (K-40) ([A₀] = 100%).

[A] is the remaining concentration of (K-40) ([A] = 6.25%).

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = ln((100%)/( 6.25%))

∴ (5.54 x 10⁻¹⁰ year⁻¹)(t) = 2.77.

∴ t = 2.77/(5.54 x 10⁻¹⁰ year⁻¹) = 5.0 x 10⁹ years.

8 0

3 years ago

Read 2 more answers

Suppose you have 81.8 g of Cd. How many Cd atoms are present? Report your answer in scientific notation using the format of 6.02

Lostsunrise [7]

Answer:

1. number of Cd atoms = 4.384 × 10²³

2. number of moles of Ti = 0.817 mol

3. molar mass of C₇H₆O₃ = 138.077 g/mol

4. moles of Sildenafil = 0.249 mmol

Explanation:

1. given data

mass of Cd = 81.8 g

number of Cd atoms = ?

molar mass of Cd = 112.411 g/mol

Solution

1st we find out the number of moles of Cd

number of moles = mass / molar mass

number of moles of Cd = 81.8 g / 112.411 g.mol⁻

number of moles of Cd = 0.728 mol

Now we find out the number of Cd atoms

number of atoms = moles × 6.022 × 10²³

number of Cd atoms = 0.728 (6.022 × 10²³)

number of Cd atoms = 4.384 × 10²³

2. Given data

Mass of Ti = 39.1 g

moles = ?

molar mass Ti = 47.867 g/mol

Solution

Now we will find out the number of moles

number of moles = mass / molar mass

number of moles of Ti = 39.1 g / 47.867 g.mol⁻

number of moles of Ti = 0.817 mol

3. Given data

chemical formula of Salicylic acid = C₇H₆O₃

molar mass of salicylic acid = ?

molar mass of carbon = 12.011

molar mass of hydrogen = 1.008

molar mass of oxygen = 15.999

molar mass of molecule = number of moles of atoms × molar mass of atoms

molar mass of C₇H₆O₃ = 7 (12.011) + 6 (1) + 3 (16)

molar mass of C₇H₆O₃ = 84.077 g + 6 g + 48 g

molar mass of C₇H₆O₃ = 138.077 g/mol

4. Given data

molar mass of Sildenafil = 474.5828 g/mol

mass given = 0.118 g

moles of Sildenafil = ?

Solution

Now we find out the number of moles

number of moles = mass / molar mass

number of moles Sildenafil = 0.118 g / 474.5828 g/mol

number of moles Sildenafil = 0.0003 mol

as this is very small amount we convert it into millimoles

millimoles = moles × 1000

millimoles of Sildenafil = 0.000249 × 1000

millimoles of Sildenafil = 0.249 mmol

4 0

2 years ago

The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water.

avanturin [10]

a. 0.137

b. 0.0274

c. 1.5892 g

d. 0.1781

e. 5.6992 g

<h3>Further explanation</h3>

Given

Reaction

2 C4H10 + 13O2 -------> 8CO2 + 10H2O

2.46 g of water

Required

moles and mass

Solution

a. moles of water :

2.46 g : 18 g/mol = 0.137

b. moles of butane :

= 2/10 x mol water

= 2/10 x 0.137

= 0.0274

c. mass of butane :

= 0.0274 x 58 g/mol

= 1.5892 g

d. moles of oxygen :

= 13/2 x mol butane

= 13/2 x 0.0274

= 0.1781

e. mass of oxygen :

= 0.1781 x 32 g/mol

= 5.6992 g

6 0

3 years ago

You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.200 M sodium benzoate

bekas [8.4K]

Answer : The volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

Explanation :

Let the volume of sodium benzoate (salt) be, x

So, the volume of benzoic acid (acid) will be, (100 - x)

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$4.00=4.20+\log \left(\frac{(\frac{0.200x}{100})}{(\frac{0.100(100-x)}{100})}\right)$

x = 29.0

The volume of sodium benzoate = x = 29.0 mL

The volume of benzoic acid (acid) = (100 - x) = (100 - 29.0) = 71 mL

Thus, the volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.

5 0

2 years ago

What is the cost of carbon in gram

Morgarella [4.7K]

Carbon is 12 grams per mole

6 0

2 years ago