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3 years ago

When you dissolve solid sugar into water, this new solution will taste sweet. the sugar went through a chemical change?

1 answer:

7 0

Basically the sugar breaks down until it’s eventually evaporated and spreads to different parts of the water solution and while it’s spreading the chemicals and the flavors in the sugar are going into the molecules and atoms in the water and mixing to make the water solution taste sweet just like sugar. So i would say it would be a physical change not a chemical change.

You might be interested in

In a chemical reaction how does the mass of the products compare with the mass of the reaction a greater thenb less thenc equal

Neko [114]

D.) It depends cuz no yeild is 100%..I mean side reactions also occur in most of the reactions. So mass of the reactant is not equal to the mass of the product. Hope it helps

5 0

3 years ago

PLEASE HELP

neonofarm [45]

Answer:

The answer to your question is 0.10 M

Explanation:

Data

Molarity = ?

mass of Sucrose = 125 g

volume = 3.5 l

Formula

Molarity = moles / volume

Process

1.- Calculate the molar mass of sucrose

C₁₂H₂₂O₁₁ = (12 x 12) + (1 x 22) + (16 x 11)

= 144 + 22 + 176

= 342 g

2.- Convert the mass of sucrose to moles

342 g of sucrose ------------------- 1 mol

125 g of sucrose -------------------- x

x = (125 x 1) / 342

x = 0.365 moles

3.- Calculate the molarity

Molarity = 0.365 / 3.5

4.- Result

Molarity = 0.10

5 0

3 years ago

Write a ground state electron configuration for each neutral atom

Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0

2 years ago

When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$

And the resulting percent yield:

$Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%$

Regards!

4 0

3 years ago

how many oxygen atoms are there in six dinitrogen monoxide molecules? express your answer as an integer

stepan [7]

Answer:

6 oxygen atoms

Explanation:

Step 1: Given data

Number of dinitrogen monoxide molecules (N₂O): 6

Number of oxygen atoms (O): ?

Step 2: Calculate the appropriate ratio

The ratio of dinitrogen monoxide molecules to oxygen atoms is 1:1.

Step 3: Use the ratio to calculate the number of oxygen atoms

6 molecule N₂O × (1 atom O/1 molecule N₂O): 6 atom O

7 0

3 years ago