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Yuki888 [10]
3 years ago
10

What is the molarity of a solution made with 64 grams of sodium hydroxide in 4 liters of water

Chemistry
1 answer:
ehidna [41]3 years ago
7 0

Answer:

The molarity of the solution: 0,27M

Explanation:

First , we calculate the weight of 1 mol of NaCl:

Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol

58,5 g---1 mol NaCl

64 g--------x= (64 g x1 mol NaCl)/58,5 g= 1, 09 mol NaCl

A solution molar--> moles of solute in 1 L of solution:

4 L-----1,09 mol NaCl

1L----x0( 1L x1,09 mol NaCl)/4L =0,27moles NaCl--->0,27M

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Which is an element hydrogen , air , carbon dioxide , water​
Basile [38]

Answer:

Hydrogen is an element

Explanation:

Hydrogen is an element with only hydrogen atoms, whereas air, carbon dioxide, and water are all made up of multiple elements with different types of atoms.

8 0
4 years ago
Read 2 more answers
scandium47 has a half-life of 35s. suppose you have a 45g sample of scadium 47 how much of the sample remains unchanged after 14
Mars2501 [29]

 The much  of the sample that would remain  unchanged  after 140 seconds is 2.813 g

Explanation

Half life  is time taken for the quantity  to reduce  to half its original value.

if the half life  for Scandium  is 35 sec, then the number  of half life in 140 seconds

=140 sec/ 35 s = 4 half life

Therefore 45 g after first half life = 45 x1/2 =22.5 g

               22.5 g after second half life = 22.5 x 1/2 =11.25 g

            11.25 g after third half life = 11.25 x 1/2 = 5.625 g

             5.625 after  fourth half life = 5.625 x 1/2 = 2.813

therefore 2.813 g  of Scandium 47 remains  unchanged.

4 0
3 years ago
9.8 x 10​^-6 regular notation
topjm [15]

Answer:

0.0000098 should be the answer

Explanation:

3 0
3 years ago
2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

3 0
3 years ago
A scientist wants to make a solution of tribasic solution phosphate, na3po4, for a laboratory experiment. How many grams of na3p
timurjin [86]

Answer:

55.75g

Explanation:

From

m/M = CV

Where

m= required mass of solute

M= molar mass of solute

C= concentration of solution

V= volume of solution=675ml

Molar mass of solute= 3(23) + 31 + 4(16)= 69+31+64=164gmol-1

Number of moles of sodium ions present= 1.5× 675/1000= 1.01 moles

Since 1 mole of Na3PO4 contains 3 moles of Na+

It implies that 1.01/3 moles of Na3PO4 are present in solution= 0.34moles

mass of Na3PO4= number of moles × molar mass= 0.34 × 164 =55.75g

3 0
4 years ago
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