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MrRissso [65]
3 years ago
9

Does salt in water evaporate  faster than sugar in water

Chemistry
2 answers:
Romashka [77]3 years ago
8 0
Yes - salt "dehydrates"water, and therefore, it would evaporate faster.
hope this helped!!
Natalka [10]3 years ago
5 0
No it does noot evaporate faster than sugar in water
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BaCl2(aq) + Na2CO3(aq) BaCO3(s) + NaCl(aq
Vitek1552 [10]
This is the equation balanced:

<span>BaCl2(aq) + Na2CO3(aq) = BaCO3(s) + 2 NaCl(aq)

Then the coefficient in front of Na Cl is 2.

Answer: 2
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7 0
3 years ago
Consider the addition of HBr to 2-pentene. Indicate the re
lana [24]

Answer:

This addition reaction yields 3-BromoPentane and 2-BromoPentane.

Explanation: The reaction is an addition reaction that follows the Markonikoff's principle engaging the electrophillic addition mechnism with electrophile having no lone pair so rearrangement of carbonation is possible. It yields two possible products.

5 0
3 years ago
1. Use the following equation:
fenix001 [56]

Answer:

7.28 mol

Explanation:

2 NaOH +  H₂SO₄ =  2 H₂O +  Na₂SO₄ -------------------(1)

mole fraction for the reaction is;

2 : 1 = 2 : 1

Number of moles of H₂SO₄ = 7.28 mol

1 mol of H₂SO₄ shall form 1 mole of Na₂SO₄

therefore,

7.28 mol of H₂SO₄ shall form 7.28 mole of Na₂SO₄

8 0
3 years ago
Read 2 more answers
A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ =  0.00375  mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

HN_3 + OH- ---> N^-_{3} + H_2O

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction =  0.025 +  0.0133 = 0.0383  L

Concentration of HN_3 = \dfrac{0.001755}{0.0383} = 0.0458 M

Concentration of N^{-}_3 = \dfrac{ 0.001995 }{0.0383} = 0.0521 M

GIven that :

Ka = 1.9 x 10^{-5}

Thus; it's pKa = 4.72

pH =4.72 +  log(\dfrac{ \ 0.0521}{0.0458})

pH =4.72 + log(1.1376)

pH =4.72 + 0.05598

pH =4.77598

pH ≅ 4.80

3 0
3 years ago
What is this equation balanced?
Brums [2.3K]
This equation is impossible. NaSO4 is non-existent. Did you mean Na2SO4?
4 0
2 years ago
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