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icang
4 months ago
14

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

Chemistry
2 answers:
Alekssandra [3K]4 months 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 [2.9K]4 months 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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A certain metal alloy is composed of 10% tin, 16% antimony, and 74% lead. If you were to have 500 g of the alloy, how many grams
Anarel [2989]
80 g. Explanation: In the metal alloy, the weight percent of antimony can be determined using the formula wt% = (mass of compound/mass of alloy)×100. Since we have 16% = (grams of antimony / grams of alloy)×100, simplifying gives us 0.16 = grams of antimony / grams of alloy, and therefore if the alloy is 500 g, it leads us to the conclusion that the weight of antimony is 80 g.
3 0
2 months ago
Two different atoms have six protons each and the same mass. However, one has a negative charge while the other has a positive c
castortr0y [3046]

The equal mass indicates that both atoms have the same number of protons and neutrons.

A positive charge signifies a difference in electron count.

Assuming the atomic number is A,

the mass number equals M.

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A positively charged atom contains A - 1 electrons [with consistent protons and mass number].

For instance: Cl- and Cl+.

8 0
3 months ago
Read 2 more answers
it takes 151 kJ/mol to break an iodine-iodine single bond. calculate the maximum wavelength of light for which an iodine-iodine
alisha [2963]

Answer:

To break a single I-I bond, the wavelength of light required is 7.92 × 10⁻⁷ m

Explanation:

The energy needed to break one mole of iodine-iodine single bonds is 151 KJ

The energy necessary to rupture one iodine-iodine bond is calculated as (151 KJ/mol) / 6.02 × 10²³/mol = 2.51 × 10⁻²² KJ

or

2.51 × 10⁻¹⁹ J

Formula:

E = hc / λ    

Where h is Planck's constant    = 6.626 × 10⁻³⁴ js

c is the speed of light = 3 × 10⁸ m/s

λ = wavelength

Solution:

E = hc / λ  

λ   = hc / E

λ   =  (6.626 × 10⁻³⁴ js × 3 × 10⁸ m/s ) / 2.51 × 10⁻¹⁹ J

λ   = 19.878 × 10⁻²⁶ j.m / 2.51 × 10⁻¹⁹ J

λ   = 7.92 × 10⁻⁷ m

6 0
3 months ago
The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. What is this distance in mil
lorasvet [2795]

Answer:

C) 1.15 × 10⁻⁷ mm

Explanation:

Step 1: Provided information

Average separation between oxygen and nitrogen atoms: 115 pm

Step 2: Change the distance to meters (SI standard unit)

Using the conversion 1 m = 10¹² pm.

115 pm × (1 m/10¹² pm) = 1.15 × 10⁻¹⁰ m

Step 3: Transform the distance to millimeters

Employing the conversion 1 m = 10³ mm.

1.15 × 10⁻¹⁰ m × (10³ mm/1 m) = 1.15 × 10⁻⁷ mm

5 0
2 months ago
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