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Bingel [31]
3 years ago
15

Write 6.3 x 10^2 in standard form. O 0.063 O 630 O 63 O 63

Chemistry
1 answer:
Leto [7]3 years ago
7 0

Answer:

it will be 630

Explanation:

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Given the following equation: 2 KCIO, +2 KCl + 30, how many grams of O, can be produced by letting 5.90 moles of KCIO, react ?
Finger [1]

Answer:

12.o of 02

Explanation:

8 0
3 years ago
The image shows sedimentary rock layers with index fossils and a fault.
GrogVix [38]

Answer:

Layer 2 and layer 9 is the same relative age since it is the same type of rock and has the same fossils.

Hope this helped!

4 0
3 years ago
When dissolved in water, a neutral molecule of calcium chloride dissociates into three ions - one calcium ion and two chloride i
Mashcka [7]

Explanation:

Calcium chloride is an ionic compound as it is formed by transfer of an electron to each chlorine atom.

So, being an ionic compound calcium chloride is able to dissociate completely into water.

Hence, the dissociation reaction will be as follows.

         CaCl_{2}(s) \rightarrow Ca^{2+}(aq) + 2Cl^{-}(aq)

Since, two electrons has been lost by single calcium atom. Therefore, calcium atom will have a charge of +2.

Thus, we can conclude that the charge on the calcium ion, in elementary units is +2.

8 0
3 years ago
A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the
tino4ka555 [31]

Answer:

2.1 × 10⁻¹ M

2.0 × 10⁻¹ m

Explanation:

Molarity

The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:

3.9 g × (1 mol/93.13 g) = 0.042 mol

The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:

M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M

Molality

The moles of solute are 0.042 mol.

The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:

200 mL × 1.05 g/mL = 210 g = 0.210 kg

The molality of aniline is:

m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m

5 0
3 years ago
Calculate the concentration (in mol/l) of 33% by weight (33 g naoh per 100 g of solution) naoh solution. (the density of the 33%
Shtirlitz [24]
Molarity is measured in moles per Liter. If there are 1.35 g/mL, find out how many grams there are in a liter of solution.

If there are 1000 mL in one liter, we can multiply by 1000 to get g/L

1.35 g/mL x 1 Liter/1000 mL = 1350 g per Liter of solution

By weight, the NaOH is 33% or .33

1350 g x .33 = 445.5 g of NaOH

Molar mass of NaOH is 39.997 g

445.5 g  x 1 mol NaOH/39.997 g = 11.13833538 moles per Liter

Rounded to significant figures, the answer is 11 mol/L NaOH


7 0
3 years ago
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