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a_sh-v [17]
3 years ago
15

A sample contains 0.1895 g of salt per 50 g of seawater. What is its salinity?

Chemistry
1 answer:
Naily [24]3 years ago
7 0
Use the rule of 3 simple
50 g mean 100% 0,1895 mean x% cross multiplie and calcule the x value x = 0,1895*100/50 = ?%
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Right answers only will be reported if not right and no lniks!!!!!!!!!!!!!!!!!
Tasya [4]

Answer:

tgratreqw

Explanation:

3 0
3 years ago
a normal hemoglobin concentration in the blood is 15g/100ml of blood how many kilogram of hemoglobin are in a person who has 5.5
jok3333 [9.3K]
Using stoichiometry:

5.5 L of blood x (1000 mL/1L) x (15 g/100 mL) x (1 kg/1000 g) = 0.825 kg
4 0
3 years ago
Read 2 more answers
Calculate the mole fraction of each component in a solution with 6.87 g of sodium chloride (NaCl) dissolved in 65.2 g of water.
zheka24 [161]

Answer:

mole fraction of NaCl = 0.03145.

mole fraction of water = 0.9686.

Explanation:

  • Mole fraction is an expression of the concentration of a solution or mixture.
  • It is equal to the moles of one component divided by the total moles in the solution or mixture.
  • The summation of mole fraction of all mixture components = 1.

mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles).

<em>no. of moles of NaCl = mass/molar mass </em>= (6.87 g)/(58.44 g/mol) = 0.1176 mol.

<em>no. of moles of water = mass/molar mass</em> = (65.2 g)/(18.0 g/mol) = <em>3.622 mol.</em>

<em></em>

∴ mole fraction of NaCl = (no. of moles of NaCl) / (total no. of moles) = (0.1176 mol)/(0.1176 mol + 3.622 mol) = 0.03145.

<em>∵ mole fraction of NaCl + mole fraction of water = 1.0.</em>

∴ mole fraction of water = 1.0 - mole fraction of NaCl = 1.0 - 0.03145 = 0.9686.

7 0
3 years ago
Why are 1-chlorobutane and 2-chlorobutane structural isomers?.
Nutka1998 [239]

Answer: they both have the same molecular formula but different structural formulae

Explanation:

hope this helps :)

6 0
2 years ago
A compound containing Na, C, and O is found to have 1.06 mol Na, 0.528 mol C, and 1.59 mol O. What is the empirical formula of t
Effectus [21]

Given :

Moles of Na : 1.06

Moles of C : 0.528

Moles of O : 1.59

To Find :

The empirical formula of the compound.

Solution :

Dividing moles of each atom with the smallest one i.e 0.528 .

So,

Na : 1.06/0.528 = 2.007 ≈ 2

C : 0.528/0.528 = 1

O : 1.59/0.528 = 3.011 ≈ 3

Rounding all them to nearest integer, we will get the number of each atom in the empirical formula.

So, empirical formula is Na_2CO_3 .

Hence, this is the required solution.

5 0
3 years ago
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