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3 years ago

How do isotopes of a givin element similar

2 answers:

6 0

Different isotopes of an element have the same number of protons in the nucleus, giving them the same atomic number, but a different number of neutrons giving each elemental isotope a different atomic weight.

BlackZzzverrR [31]3 years ago

5 0

Answer:

All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons.

Explanation:

Hope this <em><u>Helped!</u></em> :D

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An atom has three shells with 3 electrons in the last shell. This atom could be.....

Oksana_A [137]

Since the total amount of valence electrons is 3, it is in group 13 in the periodic table..therefore, it is specified as Boron.✅

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2 years ago

Based on the article "Will the real atomic model please stand up?,” describe one major change that occurred in the development o

nydimaria [60]

I don't know this article, but I do know some major changes: first, the change from the plum pudding model (no nucleus, just electrons) to the gold foil experiment, which had Rutherford shoot alpha particles at a sheet of gold only to find them rebounding, proving the existence of a positively charged mass, i.e a nucleus, in the atom. However, this changed again when Bohr realized that the negatively charged electrons should be attracted to the positively charged center, so that there must be something else inside the nucleus.

3 0

2 years ago

Read 2 more answers

How many moles of KOH are required to produce 4.79 g K3PO4 according to the following reaction? 3KOH + H3PO4 -----> K3PO4 + 3

8_murik_8 [283]

Answer:

0.677 moles

Explanation:

Take the atomic mass of K = 39.1, O =16.0, P = 31.0

no. of moles = mass / molar mass

no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)

= 0.02256 mol

From the equation, the mole ratio of KOH : K3PO4 = 3 :1,

meaning every 3 moles of KOH used, produces 1 mole of K3PO4.

So, using this ratio, let the no. of moles of KOH required to be y.

$\frac{3}{1} =\frac{y}{0.02256} \\$

y = 0.02256 x3

y = 0.0677 mol

If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.

5 0

3 years ago

How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium

Pepsi [2]

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as $\text{Pb}^{2+}$ ions. The formula for a nitrate ion is ${\text{NO}_3}^{-}$ . The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each $\text{Pb}^{2+}$ ion in lead nitrate would pair up with two ${\text{NO}_3}^{-}$ ions. The formula for lead nitrate will be $\text{Pb}({\text{NO}_3})_2$ . Each formula unit of lead nitrate will contain one $\text{Pb}^{2+}$ ion and two ${\text{NO}_3}^{-}$ ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as $\text{Cr}^{3+}$ . Similarly, each $\text{Cr}^{3+}$ will pair up with three ${\text{NO}_3}^{-}$ ions. The formula for chromium (III) nitrate will be $\text{Cr}(\text{NO}_3})_3$ . Each formula unit of chromium (III) nitrate will contain one ${\text{NO}_3}^{-}$ ion and three ${\text{NO}_3}^{-}$ ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of ${\text{NO}_3}^{-}$ ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

7 0

3 years ago

The partial pressure of N2 in the air is 593 mm Hg at 1 atm. What is the partial pressure of N2 in a bubble of air a scuba diver

Evgen [1.6K]

Answer: Partial pressure of $N_{2}$ at a depth of 132 ft below sea level is 2964 mm Hg.