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3 years ago

CaCO3(s)+2H*(aq) →Ca2+ (aq)+H2001+CO262)

Chemistry

1 answer:

Effectus [21]3 years ago

8 0

Answer:CO2(g) will be formed at a faster rate in experiment 2 because more H+ particles can react per unit time

Explanation:

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A gas occupies a volume of 4.50 L at 44 kPa. What would be the new volume at 50 kPa?

oksano4ka [1.4K]

Explanation:

P1= 44 kpa

P2= 50 kpa

V1= 4.50 L

V2= ?

P1 V1= P2 V2

44 × 4.50 = 50 × V2

198= 50 × V2

V2 = 198/ 50

V2= 3.96 L "the new volume"

5 0

3 years ago

Water vapor condenses into a glass of water. The water is then placed in a freezer and changes into ice. What remains consistent

PolarNik [594]

Chemical properties

8 0

3 years ago

Does beryllium (Be) or Sodium (Na) have the same electron arrangement as Magnesium (Mg)? Why?

Finger [1]

Answer:

Na has the most similar configuration.

Explanation:

Na electron configuration: 1s²2s²2p⁶3s¹ or [Ne] 3s₁

Mg electron configuration: 1s²2s²2p⁶3s² or [Ne] 3s²

Be electron configuration: 1s²2s² or [He] 2s²

This is because Na and Mg are right next to each other in the same period (horizontal).

3 0

3 years ago

Hellllppppp lons Worksheet

Iteru [2.4K]

The metals will lose electrons while the non metals will gain electrons in order to attain octet structure.

An ion can be cation (positively charged) or anion (negatively charged).

Cations attain octet structure (8) by losing electron(s) while anions become stable or attains octet structure (8) by gaining electron(s).

The remaining elements are completed as follows to attain octet structure;

Element--valence electron--electrons to gain--electrons to lose--ion formed

O ------------ 6 ---------------------- 2 ------------------------ none -------------- $O^{2-}$

Ca -------- 2 ----------------------- none ---------------------- 2 ------------------ $Ca^{2+}$

Br ----------- 7 --------------------- 1 ------------------------ none --------------- $Br^{-}$

S ------------ 6 ----------------------- 2 ------------------------ none --------------- $S^{2-}$

Cl ------------ 7 ----------------------- 1 ------------------------ none ---------------- $Cl^{-}$

K -------------- 1 ----------------------- none ----------------------- 1 ------------------ $K^{+}$

Mg ------------ 2 ---------------------- none ---------------------- 2 ---------------- $Mg^{2+}$

Be ------------- 2 ---------------------- none ---------------------- 2 ---------------- $Be^{2+}$

Learn more here: brainly.com/question/21089350

8 0

3 years ago

A gas cylinder of volume 5.00 l contains 1.00 g of ar and 0.500 g of ne. the temperature is 275 k. find the partial pressure of

amid [387]

11.3 kPa The ideal gas law is PV = nRT where P = Pressure V = Volume n = number of moles R = Ideal gas constant (8.3144598 L*kPa/(K*mol) ) T = Absolute temperature We have everything except moles and volume. But we can calculate moles by starting with the atomic weight of argon and neon. Atomic weight argon = 39.948 Atomic weight neon = 20.1797 Moles Ar = 1.00 g / 39.948 g/mol = 0.025032542 mol Moles Ne = 0.500 g / 20.1797 g/mol = 0.024777375 mol Total moles gas particles = 0.025032542 mol + 0.024777375 mol = 0.049809918 mol Now take the ideal gas equation and solve for P, then substitute known values and solve. PV = nRT P = nRT/V P = 0.049809918 mol * 8.3144598 L*kPa/(K*mol) * 275 K/5.00 L P = 113.8892033 L*kPa / 5.00 L P = 22.77784066 kPa Now let's determine the percent of pressure provided by neon by calculating the percentage of neon atoms. Divide the number of moles of neon by the total number of moles. 0.024777375 mol / 0.049809918 mol = 0.497438592 Now multiply by the pressure 0.497438592 * 22.77784066 kPa = 11.33057699 kPa Round the result to 3 significant figures, giving 11.3 kPa

8 0

3 years ago