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kaheart [24]
2 years ago
6

What is the concentration in milligrams per milliter of a solution containing 23.5 meq sodium chlorise per milliliter? mw nacl =

58.5? 2
Chemistry
1 answer:
ValentinkaMS [17]2 years ago
7 0

1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.

It is calculated in mg/ml.

The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.

Formula for calculating concentration in mg/ml is

Conc. (mg/ml) = M(eq) /ml ×  Molecular weight / Valency

Given

M(eq) NaCl/ ml = 23.5

Molecular weight pf NaCl = 58.5 g/mol

Valency = 1

Putting the values into the formula

Conc. (mg/ml) = 23.5 ×58.5/1

                       = 1374.75 mg/ml

Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.

Learn more about Concentration here brainly.com/question/14500335

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What does the roman numeral stand for in copper(1) oxide should it not be copper(II) oxide
photoshop1234 [79]

Answer:

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

Explanation:

Roman numeral is used to indicate the oxidation number of an element in a compound.

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

This can be seen from the following illustration:

copper(I) oxide => Cu₂O

Oxidation number of O = –2

Oxidation number of Cu₂O = 0

Oxidation number of Cu =?

Cu₂O = 0

2Cu + O = 0

2Cu – 2 = 0

Collect like terms

2Cu = 0 + 2

2Cu = 2

Divide both side by 2

Cu = 2/2

Cu = 1

Thus, we can see that the oxidation number of Cu in Cu₂O is 1. Hence the name of Cu₂O is copper(I) oxide indicating that the oxidation number of of copper (Cu) in the compound is 1.

For copper(II) oxide, we shall determine the oxidation number of Cu. This can be obtained as follow:

copper(II) oxide, CuO => CuO

Oxidation number of O = –2

Oxidation number of CuO = 0

Oxidation number of Cu =?

CuO = 0

Cu + O = 0

Cu – 2 = 0

Collect like terms

Cu = 0 + 2

Cu = 2

Thus, the oxidation number of Cu in CuO is 2. Hence the name of CuO is copper(II) oxide indicating that the oxidation number of of copper (Cu) in the compound is 2.

From the above illustrations,

We can see that the roman numeral in both copper(I) oxide, Cu₂O and copper(II) oxide, CuO are different because the oxidation number of Cu in both cases are different.

3 0
3 years ago
A 10.0 gram sample of Fe contains how many miles of Fe
Dafna1 [17]

Answer:

1.7857 moles

Explanation:

moles=mass/Mr

10/56=1.7857

moles of iron =1.7857

4 0
2 years ago
What are the atomic number and mass number of this atom
lina2011 [118]
No diagram is given therefore I cannot answer however, use this for reference.
The atomic mass of an atom would be the number of protons PLUS the number of neutrons. Because this atom shown has 6 protons and 6 neutrons, they add up to 12. In this case, the atomic mass of this atom is 12. Atomic number is the number of protons in an atom, in this case it is 6. The atomic number is 6.

5 0
3 years ago
When the fuel mixture contained in a 1.49 l tank, stored at 750 mmhg and 298 k, undergoes complete combustion, how much heat is
Ket [755]
Get to know first how many moles in the gas:n = pV/RT= (1.013*10^5*750/760) Pa *1.49*10^-3 m^3/(8.314 J/(molK)*298) n = 0.0601 moles. 
The combustion energies  are  889 kJ/mol (methane) and 2 220 kJ (propane) x = moles methane, y = moles propane 
x*889 + y*2220 = 778 x + y = 0.0601----------- x = 0.267784 moles = 0.267784*100/0.0601 = 44.6 % y = 0.243216 moles = 0.243216*100/0.0601 = 55.4 %
4 0
3 years ago
A mixture of caco3 and (nh4)2co3 is 60.7 % co3 by mass. part a find the mass percent of caco3 in the mixture.
blagie [28]

The mixture contains:

CaCO3 + (NH4)2CO3 in which the amount of carbonate CO3 = 60.7% by mass

Let, the total mass = 100 grams

Mass of CaCO3 = x grams

Mass of (NH4)2CO3 = y grams

Thus,       x + y = 100 ------------(1)

Mass of CO3 = 60.7% = 60.7 g

Molar mass of CO3 = 60 g/mol

Total # moles of CO3 = 60.7 g/60 g.mol-1 = 1.012 moles

The total moles of CO3 comes from CaCO3 and (NH4)2CO3. Therefore,

moles CaCO3 + moles (NH4)2CO3 = 1.012

mass CaCO3/molar mass CaCO3 + mass (NH4)2 CO3/molar mass = 1.012

x/100 + y/96 = 1.012---------(2)

based on equation 1 we can write: y = 100-x

x/100 + (100-x)/96 = 1.012

x = 71.2 g

Mass of CaCO3 = 71.2 g



7 0
3 years ago
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