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icang
9 days ago
14

How would you convert 500cc of 2M H2SO4 into g/l?​

Chemistry
2 answers:
Alekssandra [968]9 days ago
7 0

Answer:

Count of replaceable H

+

ions present in H

2

SO

4

=n=2

Normality =n× Molarity=2×2=4 N

KiRa [971]9 days ago
0 0

gm/l = molecular weight * molarity


Thus, the molarity provided for H2SO4 is 2M. The molecular weight is known to be 98 amu;

Therefore, gm/l = 98 * 2 = 196

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Explain why CaCl2 is likely to have properties similar to those of CaBr2
Alekssandra [968]

Answer:

Both CaCl2 and CaBr2 consist of elements (bromine and chlorine) from the same group (group 7).

Explanation:

In the periodic table, elements are arranged into groups based on their valence electron count in the outermost shell. Elements in the same group, which possess a similar number of valence electrons, typically exhibit similar chemical behaviors.

Chlorine and Bromine in CaCl2 and CaBr2 belong to group 7, known as HALOGENS, characterized by having 7 valence electrons in their outer shell.

The similarity in properties between CaCl2 and CaBr2 arises because both contain Chlorine and Bromine, leading to analogous reactions and behaviors when interacting with other compounds.

5 0
15 days ago
Compaction is most significant as a lithification process for sedimentary rocks composed of sand-sized particles. True False
VMariaS [1037]

Response:

FALSE

Rationale:

Sedimentary rocks are defined as rocks formed through the processes of compaction and cementation. Initially, sediments derived from various locations must accumulate. Over time, these deposited sediments undergo substantial compaction due to the weight of the layers above. This process converts loose sediments into solid rock. This is the process through which sedimentary rocks, comprised of sand-sized particles, are formed. For instance, examples include Shale, Sandstone, and Mudstone.

Thus, both compaction and cementation are crucial in the formation of sedimentary rocks.

Therefore, the statement above is False.

8 0
5 days ago
At what temperature would the volume of a gas be 0.550 L if it had a volume of 0.432 L at –20.0 o C?
castortr0y [923]
To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.

The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L

Rearranging for T1:
T1 = (V1 × T2) / V2

Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C
8 0
13 days ago
Read 2 more answers
A 6.1-kg solid sphere, made of metal whose density is 2600 kg/m3, is suspended by a cord. When the sphere is immersed in a liqui
castortr0y [923]

Answer:

The calculated density of the liquid is 1470.43 kg/m³.

Explanation:

Given:

Mass of the solid sphere (m) = 6.1 kg

Density of the metal = 2600 kg/m³

To find the volume of the liquid:

Volume(V)=\frac{Mass(m)}{Density (\rho)}

Volume of the sphere can be calculated as 6.1 kg / 2600 kg/m³ = 0.002346 m³.

According to Archimedes' principle, the volume of water displaced is equal to the volume of the sphere.

Volume displaced = 0.002346 m³

The buoyant force formula is:\rho\times gV

Where:

\rho is the fluid's density,

g represents the acceleration due to gravity,

V indicates the volume displaced.

Referencing the free-body diagram of the sphere shown in the image:

mg=\rho\times gV+T

Acceleration due to gravity = 9.81 ms⁻²

Tension force = 26 N

Using these in the equation to ascertain the liquid density yields:

6.1\times 9.81=\rho\times 9.81\times 0.002346+26

33.841=\rho\times 9.81\times 0.002346

\rho=\frac{33.841}{9.81\times 0.002346}

\rho=1470.43 kgm^3

Thus, the density of the liquid = 1470.43 kg/m³

6 0
8 days ago
Why the gross reading is needed when doing the titration? ​
Anarel [852]

Answer:

The response is provided below.

Explanation:

Numerous aspects can influence the actual results of titration. These factors vary from human error to misjudging measurements, a researcher's interpretation of color changes, and improper techniques during the experimental procedure.

Thus, to mitigate these errors, researchers must engage thoroughly throughout experimentation, and employing gross readings can assist in reducing mistakes when determining the final titre value.

7 0
7 days ago
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