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3 years ago

Which metal is most likely to form more than one kind of positively charged ion

Chemistry

2 answers:

Aleks [24]3 years ago

6 0

Answer:

Nickel (Ni) is most likely to form more than one kind of positively charged ion as it is a transition metal and all of the ther metals have a fixed ion charge.

Na becomes Na+

Ba becomes Ba +2

Mg becomes Mg+2

Explanation:

please mark brailiest

polet [3.4K]3 years ago

6 0

Answer:

metal atoms lose electrons to form positivity charged ions. non metal atoms gain electrons to form negatively charged ions

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Practice Explaining Heat Flow

zloy xaker [14]

Answer:

The right choice is "burning gas → pan → water → eggs"

Explanation:

Heat will flow from the body of higher temperature to the cooler body.

So, heat will firstly flow from burning gas to pan by conduction then to the water in contact with pan then to the egg present inside the water.

<u>So, the right choice is :</u>

"burning gas → pan → water → eggs"

6 0

3 years ago

Read 2 more answers

What are the products of the combustion of a hydrocarbon?

Oxana [17]

Answer:

I think carbon and hydrogen

7 0

3 years ago

Read 2 more answers

calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

5 0

3 years ago

How many moles of oxygen will occupy a volume of 2.50 liters?

loris [4]

Data:
n (number of mols) = ?
V (volume) = 2.50 Liters
If:
1 L → 1000 g
2.50 L → y

y = 1000*2.50 = 2500 g

Therefore:
m (mass) = 2500 g

Now:
Molar Mass (MM) of oxygen = 16 g/mol

Formula:
$n = \frac{m}{MM}$

Solving:
$n = \frac{m}{MM}$
$n = \frac{2500\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol}$
$\boxed{\boxed{n = 156.25\:mols}}\end{array}}\qquad\quad\checkmark$

7 0

3 years ago

which atoms attains a stable valence electron configuration by bonding with another atom 1) neon 2) radon 3) helium 4) hyrdogen

Sergeu [11.5K]

Well, when an atom attains a stable valence electron, it means that the outer electrons are complete and so cannot attain any more electrons. For the first shell, it is complete when it has 2 electrons, the second shell is complete when it has 8 electrons, all the other shells also have a particular number when complete. Anyway, i believe the answer is HYDROGEN because when HYDROGEN combines with another atom of HYDROGEN, the outer shell is completed. This is because HYDROGEN has only 1 electron. If the two HYDROGENS, which both have 1 electron combine, they make the electrons 2, which is complete for the first shell, HYDROGEN ends in the first shell. Since the electrons become 2, the shell is at stable valence. In all the other options, this happens;
NEON- It has 10 electrons, 2 in the first shell and 8 in the second. So the the shells are already complete, so it can't bond with any thing, which is completely against the question.
RADON- Radon has 86 electrons.
HELIUM- Helium has 2 electrons, so the shell is already full, and cannot bond, so it goes against the question. The question says BY BONDING.
So the answer is definitely 4) HYDROGEN
Hope i helped. Have a nice day, by the way, i'm very sure it's hydrogen.

3 0

3 years ago