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2 years ago

An atom of chromium has 24 protons, 28 neutrons, and 24 electrons. A chromium atom has_____Subatomic particles in the necleus

Chemistry

1 answer:

navik [9.2K]2 years ago

5 0

3 subatomic particles in the nucleus
which is proton(24),neutron (28) and electrons(24)

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Atomic number for bromine

Yuliya22 [10]

Atomic number for bromine is 35

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2 years ago

Bronsted-Lowry bases are proton<br> acceptors<br> donors<br> reducers

Scrat [10]

Answer:

acceptor

Explanation:

3 0

2 years ago

What is the formula for manganese (ii) fluoride decahydrate? (you may use a * to represent the dot in the formula.)?

Rasek [7]

These types of molecules are called hydrates. They have a certain number of moles attached to the salt. Their characteristic is being hygroscopic. That means that when they are exposed to air, they readily solvate.

The formula for Manganese Fluoride Decahydrate will involve the formula Mn, F and H₂O. In ionic form, Manganese is Mn⁺² while fluoride is in F⁻. When they are brought together, their superscripts are 'cross-multiplied' and becomes their respective subscripts. The compound becomes MnF₂. Then, we add the decahydrate which means 10 moles of H₂O. Hence, the formula for Manganese Fluoride Decahydrate is MnF₂*10H₂O.

4 0

2 years ago

Read 2 more answers

The gas phase decomposition of sulfuryl chloride at 600 K SO2Cl2(g)SO2(g) + Cl2(g) is first order in SO2Cl2. During one experime

krok68 [10]

Answer: The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample = 559 min

a = let initial amount of the reactant = $2.83\times 10^{-3}$

a - x = amount left after decay process = $3.06\times 10^{-4}$

$559min=\frac{2.303}{k}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=\frac{2.303}{559}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=3.96\times 10^{-3}min^{-1}$

The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

6 0

2 years ago

How many grams of KCIO3 are needed to produce 5.00 of O2 at STP?

Arlecino [84]

2KClO₃ → 2KCl + 3O₂

mole ratio of KClO₃ to O₂ is 2 : 3

∴ if moles of O₂ = 5 mol

then moles of KClO₃ = $\frac{5 mol * 2}{3}$

= 3.33 mol

Mass of KClO₃ needed = mol of KClO₃ × molar mass of KClO₃

= 3.33 mol × ((39 × 1) + (35.5 × 1) + (16 × 3) g/mol

= 407.93 g

6 0

2 years ago