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Alexus [3.1K]
2 years ago
8

How many pounds of gold are present in 1 cubic mile of seawater?

Chemistry
1 answer:
atroni [7]2 years ago
8 0

Answer:

38 pounds of gold in cubic mile of seawater.

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What is the equilibrium vapor pressure of a liquid and how is it measured
weqwewe [10]
The equilibrium vapor pressure of a liquid is the pressure exerted by a vapor on the liquid at a given temperature. It is measured in atm.
8 0
3 years ago
A 85.4 lb child has a Streptococcus infection. Amoxicillin is prescribed at a dosage of 45 mg per kg of body weight per day give
bixtya [17]

The amount of Amoxicillin dose given to the 85.4 lb child daily is determined as 1,743.3 mg.

<h3>What is the amount of Amoxicillin dose given to the child?</h3>

The amount of  Amoxicillin  dose given to the child is calculated as follows;

amount of Amoxicillin dose = weight of the child x dosage prescribed

<h3>What is the weight of the child in pounds (lb) </h3>

The weight of the child in pounds (lb) is calculated as follows;

1 lb = 0.453592 kg

85.4 lb = ?

= 85.4 x 0.453592 kg

= 38.74 kg

amount of Amoxicillin dose = 38.74 kg x 45 mg/kg

amount of Amoxicillin dose = 1,743.3 mg

Thus, the amount of Amoxicillin dose given to the 85.4 lb child daily is determined as 1,743.3 mg.

Learn more about amount of dose here: brainly.com/question/11185154

#SPJ1

The complete question is below:

A 85.4 lb child has a Streptococcus infection. Amoxicillin is prescribed at a dosage of 45 mg per kg of body weight per day given b.i.d. Calculate the daily dose of the child.

7 0
1 year ago
A gas has a volume of 25.0 mL when under a pressure of 525 mmHg. What is the new pressure when the volume has been increased to
Fudgin [204]

Answer:

152.26 mmHg

Explanation:

pv=p'v'

525× 25=p'×86.2

p'=525×25÷ 86.2

p'=152.26

5 0
3 years ago
A scientist measures the standard enthalpy change for the following reaction to be 67.9 kJ:
bekas [8.4K]

Answer: The standard enthalpy of formation  of H_2O(g) is  -252.1 kJ/mol.

Explanation:

The balanced chemical reaction is,

Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]

Putting the values we get :

\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]

67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)]

H_f{H_2O}=-252.1kJ/mol

Thus standard enthalpy of formation  of H_2O(g) is  -252.1 kJ/mol.

3 0
3 years ago
At which pesan pressure lowest The image shows a representation of mountains of various heights, numbered 1, 2, and 3 The plami
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1 is the lowest pressure.........

4 0
2 years ago
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