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VMariaS [17]
3 years ago
13

What is the formula for the compound formed by lead(lV) ions and chromate ions

Chemistry
1 answer:
nexus9112 [7]3 years ago
8 0

Answer:

Lead (IV) Chromate: Pb(CrO4)2

Explanation:

You might be interested in
A strontium-90 atom that has a lost two electrons has __________ protons, __________ neutrons, and __________ electrons.
MAXImum [283]
<span><em>Answer:</em>
A strontium-90 atom that has a lost two electrons has <u>38</u> protons, <u>52</u> neutrons, and <u>36</u> electrons.

<em>Explanation:
</em>Atomic number<em> of </em>Strontium (Sr) is 38. 
<em>Atomic number = number of protons 
</em>Hence, Strontium has 38 protons.

If the element is in neutral state,
                number of protons = number of electrons. 
Then, neutral Strontium atom should have 38 electrons. 
But the question says Sr has lost 2 electrons. Hence, number of electrons should be 38 - 2 = 36.

Mass number = number of protons + number of neutrons.

The given mass number is 90. Hence, number of neutrons should be 90 - 38 = 52.</span>
4 0
3 years ago
An acidic fog in pasadena was found to have a ph of 2.50. which expression represents this ph measurement?
joja [24]

Supposing a temperature of 25 degrees and supposing that all activity coefficients are 1 

pH = -log[H+] 
pOH = -log[OH-] 
pH + pOH = 14 

Thus a pH of 2.50 would mean that the [H+], the concentration of the hydrogen ion, would be 10^(-2.50) 

pH + pOH = 14 
pOH = 14 - pH = 14 - 2.5 = 11.5 
MOH- levels would be coordinated with pOH 
pOH = -log[OH-] ==> [OH-] = [MOH-] = 10^-pOH = 10^-11.5 = 3.2 x 10^-12 

Therefore, MOH¯ = 3.2 × 10¯12 M 

5 0
3 years ago
Can you Convert 75°F to °C
Leya [2.2K]

Answer: 23.8889

Explanation:

(75°F − 32) × 5/9 = 23.889°C

5 0
3 years ago
5. _______ is the combining capacity of atom with other atoms to form a molecule or a compound .​
Nitella [24]

Answer:

valancy

Explanation:

7 0
3 years ago
The PH solution of NH3 of 0.950molar solution is 11.612. find the Kb​
Vaselesa [24]

Answer:

1.8 x 10⁻⁵

Explanation:

 NH3(aq)  +  H2O(l)  ⇄ NH4⁺(aq)  +   OH⁻(aq)

I  0.95                              0                    0

C -x                                 +x                  +x

E 0.95-x                           x                    x

Kb= [NH₄⁺] [OH⁻] / (  NH₃) = x²/ (0.95-x )

P(OH) = 14-PH = 14-11.612 = 2.388

(OH)⁻¹ = 10⁻²°³⁸⁸ = 4.09 x 10⁻³ = x

Kb = (4.09 x 10⁻³)²/ (0.95-4.09 x 10⁻³)

= 1.8 x 10⁻⁵

8 0
3 years ago
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