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lozanna [386]
3 years ago
14

How many grams in a kilogram,how many quarts are in a pint and if how many ml of h20 = 1g

Chemistry
2 answers:
vodomira [7]3 years ago
5 0

Answer:

1 g  +  10 g

Explanation:

spayn [35]3 years ago
4 0

Answer:

there are a thousand grams in a kilo

You might be interested in
Why do farmers spray water over their crops before a frost?
Nezavi [6.7K]

Answer:

When water freezes and turns into ice, it releases latent heat. Then, the ice that builds up on the plant will insulate it from the colder surrounding air temperatures. Because of this, some growers choose to spray their crop with water before the freeze occurs.

Explanation:

5 0
3 years ago
What is the oxidizing agent in galvanic cell?
ipn [44]
The ion in the cathode that gains electrons
7 0
3 years ago
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M 

CH3COONa = 0.0076 / 0.071 = 0.1070M 

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74. 

pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

pH = 4.74 + log ( 0.1070/0.1493) 

pH = 4.74 + log 0.717 

pH = 4.74 + (-0.14) 

pH = 4.60.

7 0
3 years ago
19. What is the density of a 16.39
slavikrds [6]
Density= mass/volume

16.39g/18.00mL = 0.9105 g/mL
Make sure to use the correct number of significant figures (4)
7 0
3 years ago
Equation Given : Al^(3+) + Na3PO4 ==> 3Na^+ + AlPO4
Helga [31]

1 mols of Aluminium ion forms 1 mol aluminium phosphate

Molar mass of AlPO_4

  • 27+31+16(4)
  • 58+48
  • 106u

Moles of AlPO_4

  • 61µg/106
  • 0.000061/106
  • 5.75×10^{-7}
  • 57.5µmol

Moles of Al3+=57.5µmol

3 0
2 years ago
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