Zinc has a mass of 4.50g calculate the volume

Which of the following statements about monosaccharide structure is true?

Nana76 [90]

Answer:

The only statement about monosaccharide structure which is true is b. (Monosaccharides can be classified according to the spatial arrangement of their atoms)

Explanation:

Monosaccharides are simple sugars that are classified according to the amount of carbon atoms and based on these numbers, we can call them trioses, pentoses and hexoses. They are molecules with aldehyde (aldose) or centone (ketose) groups that have more than one alcohol function, but which do not differ in their position (OH). They do not contain N, since their general formula is Cx (H2O) x. A 6-carbon monosaccharide is called hexose, since the pentose only has 5

8 0

2 years ago

Write the electron configurations for the following ions:

Ket [755]

Answer:

Co²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Sn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zr⁴⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Ag⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

S²⁻ : 1s² 2s² 2p⁶ 3s² 3p⁶

Explanation:

Cobalt (Co): atomic number 27

The electronic configuration of Co in ground state:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷

The electronic configuration of Co in +2 oxidation state (Co²⁺) :

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷

Tin (Sn): atomic number 50

The electronic configuration of Sn in ground state:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p²

The electronic configuration of Sn in +2 oxidation state (Sn²⁺) :

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰

Zirconium (Zr): atomic number 40

The electronic configuration of Zr in ground state:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d²

The electronic configuration of Zr in +4 oxidation state (Zr⁴⁺) :

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

Silver (Ag): atomic number 47

The electronic configuration of Ag in ground state:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹ 4d¹⁰

The electronic configuration of Ag in +1 oxidation state (Ag⁺) :

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 4d¹⁰

Sulphur (S): atomic number 16

The electronic configuration of S in ground state:

1s² 2s² 2p⁶ 3s² 3p⁴

The electronic configuration of S in -2 oxidation state (S²⁻) :

1s² 2s² 2p⁶ 3s² 3p⁶

8 0

2 years ago

Read 2 more answers

A chemist mixes 75.0 g of an unknown substance at 96.5C w/ 1,150 g of water at 25.0C . if the final temperature of the system is

vichka [17]

Remember: heat lost = heat gained

When calculating heat loss or gain, remember

mass*(spec heat cap)*(change in T)

The unknown loses heat- we don't know the spec heat cap, so we'll call it x.

The water gains. I've omitted the units, but always use when solving problems on your own.

75*x*(96.5-37.1) = 1150*4.184*(37.1-25)

Now it's all set up- use algebra to get x, the spec heat cap of the unk in J/g*degC

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

3 0

2 years ago

Exactly 500 grams of ice are melted at a temperature of 32°f. (lice = 333 j/g.) calculate the change in entropy (in j/k). (give

denpristay [2]

Entropy Change is calculated by (Energy transferred) / (Temperature in kelvin)
deltaS = Q / T

Q = (mass)(latent heat of fusion)
Q = m(hfusion)
Q = (500g)(333J/g) = 166,500J

T(K) = 32 + 273.15 = 305.15K
deltaS = 166,500J / 305.15K
deltaS = 545.63 J/K

3 0

2 years ago

A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimete

N76 [4]

Answer:

The mass of water = 219.1 grams

Explanation:

Step 1: Data given

Mass of aluminium = 32.5 grams

specific heat capacity aluminium = 0.921 J/g°C

Temperature = 82.4 °C

Temperature of water = 22.3 °C

The final temperature = 24.2 °C

Step 2: Calculate the mass of water

Heat lost = heat gained

Qlost = -Qgained

Qaluminium = -Qwater

Q = m*c*ΔT

m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)

⇒with m(aluminium) = the mass of aluminium = 32.5 grams

⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C

⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C

⇒with m(water) = the mass of water = TO BE DETERMINED

⇒with c(water) = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C

32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9

-1742.1 = -7.95m

m = 219.1 grams

The mass of water = 219.1 grams

8 0

2 years ago