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salantis [7]
3 years ago
12

For the following questions protons are blue and neutrons are red

Chemistry
1 answer:
faltersainse [42]3 years ago
6 0
I can’t see the picture where’s it at
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Which tool would give the most precise measurement of 15 milliliters of water
vagabundo [1.1K]

A graduated cylinder

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6 0
2 years ago
a red box and a blue box are on the same shelf. There is more gravitational potential energy between the red box and Earth than
Anna [14]
Blue box. Gravitational potenti  al energy<span> changes into kinetic </span>energy. The equation for    gravitat ional potential energy<span> is GPE = mgh, where m i   s the mass in kilograms, g is the a   cceleration due to gravity    (9.8 on Earth), and h is the height above the ground in meters</span>
8 0
3 years ago
What is the oxidation number of Nitrogen in HNO2? <br><br> +1<br> -1<br> +3<br> -3
WITCHER [35]
I believe the correct answer from the choices listed above is the third option. The <span>oxidation number of Nitrogen in HNO2 would be +3. It is calculated as follows:

1 + x + (-2)(2) = 0
x = +3

Hope this answers the question. Have a nice day.</span>
5 0
3 years ago
If a 5¢ coin has a mass of 5.07 g and is 75.0% copper, what is the mass of copper in the coin?
ollegr [7]

Answer:

3.80g

Explanation:

5 0
3 years ago
The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo
Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is 9.57\times 10^{-10}

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)  

                            3s                  s

The expression of K_{sp} for above equation follows:

K_{sp}=(3s)^3\times s

We are given:  

s=2.44\times 10^{-3}M

Putting values in above expression, we get:

K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}

Hence, the solubility product of silver (I) phosphate is 9.57\times 10^{-10}

4 0
3 years ago
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