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Monica [59]
3 years ago
13

A 20.0-g sample of sugar is dissolved in 59.8 g of water. What is the concentration of the solution?

Chemistry
2 answers:
Romashka-Z-Leto [24]3 years ago
4 0
The correct answer to the question is 25 percent by mass
Lorico [155]3 years ago
3 0
25 percent by Mass.
Hope this helps.
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Please answer these questions
lions [1.4K]
I found number one
Atoms are not drawn to scale. Molecules of compounds have atoms of two or more different elements. For example, water (H2O) has three atoms, two hydrogen (H) atoms and one oxygen (O) atom. Methane (CH4), a common greenhouse gas, has five atoms, one of carbon (C) and four of hydrogen (H, see Fig.
7 0
3 years ago
Read 2 more answers
In what area of the titration curve for phosphoric acid (h3po4) is the molecule fully deprotonated?
Arada [10]

The titration curve should have flat regions centered around each of the

three halfway points (buffer zones) and sharp increases in pH around the

equivalence points.

Initial pH

This is determined by the most acidic of the Ka values and the initial

concentration of the acid. (Same as a monoprotic acid)

Half-way points

At each halfway point, the pH = pKa of the group you are titrating. At this point in

the titration curve, we are in a buffering region, and the curve will be relatively

flat.

Equivalence points

At each equivalence point, the pH is the average of the pKa values above and

below. At the last equivalence point (the endpoint), the pH is determined by the

Kb of the conjugate base of the weakest acid.

Plotting the titration of 100 mL of 0.10 M phosphoric acid with 1.0 M NaOH.

H3PO4 + H2O  H2PO4

-

+ H3O+

Ka1 = 7.5 x 10-3

H2PO4

-

+ H2O  HPO4

2- + H3O+

Ka2 = 6.2 x 10-8

HPO4

2- + H2O  PO4

3- + H3O+

Ka3 = 4.2 x 10-13

Plot these points and connect them to determine the titration curve of phosphoric acid. The curve should be relatively flat around each of the halfway points when we are in a buffering region.

The titration curve should have flat regions centered around each of the

three halfway points (buffer zones) and sharp increases in pH around the

equivalence points.

For more information on the titration curve click on the link below:

brainly.com/question/3130161

#SPJ4

4 0
2 years ago
What is the oxidation number of nitrogen in nitrogen gas?
andrew11 [14]
Ans: (D) 0
Hope my answer is correct:))
3 0
2 years ago
Calculate the energy released in each of the following fusion reactions. Give your answers in MeV. (a) 2H + 3H → 4He + n 17.54 C
Vikentia [17]

Answer:

a) E = 17.55 MeV

b) E = 18.99 MeV

c) E = 3.29 MeV

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) E = 4.075 MeV

Explanation:

Energy Released, E = \triangle M * 931.5

\triangle M = \sum M_{product} - \sum M_{reactant}

Mass of 1H, M_{H} = 1.007823

Mass of 2H, M_{2H} = 2.0141u

Mass of 3H, M_{3H} = 3.016 u

Mass of Helium, M_{4He} = 4.002602u

Mass of Beryllium, M_{7Be} = 7.01693 u

Mass of neutron, M_{n} = 1.008664 u

a) 2H + 3H \rightarrow 4He + n

\triangle M = (4M_{He} + M_{n} ) - (2M_{H} + 3 M_{H} )\\\triangle M = ( 4.0026 + 1.008664) - (2.0141 + 3.016 )\\\triangle M = -0.01884u

Energy released,

E = -0.01884 * 931.5\\E = -17.55 Mev

Energy released = 17.55 MeV

b) 4He + 4He \rightarrow 7Be + n

\triangle M = (M_{7Be} + M_{n} ) - (M_{4He} +  M_{4He} )\\\triangle M = ( 7.01693 + 1.008664) - (4.002602 + 4.002602 )\\\triangle M = 0.02039 u

Energy released,

E = 0.02039 * 931.5\\E = 18.99 Mev

c) 2H + 2H \rightarrow 3 He + n

\triangle M = (M_{3He} + M_{n} ) - (M_{2H} +  M_{2H} )\\\triangle M = ( 3.016 + 1.008664) - (2.0141 + 2.0141 )\\\triangle M = -0.003536 u

Energy released,

E = -0.003536 * 931.5\\E = -3.29 Mev

E = 3.29 MeV(Energy is released)

d) You can use the methods applied for the other parts to solve this, the equation is not properly written

e) 2H + 2H \rightarrow 3H + 1H

\triangle M = (M_{3H} + M_{1H} ) - (M_{2H} +  M_{2H} )\\\triangle M = ( 3.016 + 1.007825) - (2.0141 + 2.0141 )\\\triangle M = -0.00435 u

E = -0.00435 * 931.5\\E =-4.075 Mev

E = 4.075 MeV ( Energy is released)

5 0
3 years ago
Which of the following represents the electron configuration of rhodium
nordsb [41]

Answer:

i need a picture to solve

Explanation:

3 0
4 years ago
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