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3 years ago

According to the periodic table carbon has a tendency to bond covalently with

Chemistry

2 answers:

blondinia [14]3 years ago

5 0

Answer:

Carbon tends to bond covalently with Hydrogen

lisov135 [29]3 years ago

3 0

Answer:

All the carbon group atoms, having four valence electrons, form covalent bonds with nonmetal atoms; carbon and silicon cannot lose or gain electrons to form free ions, whereas germanium, tin, and lead do form metallic ions but only with two positive charges.

Explanation:

You might be interested in

What is the molecular formula of a compound with an empirical formula of C4H9 and a gram-formula mass of 114.0 g/mole?

hodyreva [135]

The formula is easy:

Gram formula mass
----------------------------
Emprical Formula

So if we plug in our own numbers,

114.0
--------
57.116

We get an answer of 1.99, which rounds to 2.
Then, we distribute based off of our empirical formula.

2(C4H9) becomes C8H18.

Our molecular formula is C8H18.
Hope I could help!

6 0

4 years ago

What mass of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution?

inna [77]

Answer:

459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution

Explanation:

Molarity is a measure of the concentration of a solute in a solution that indicates the amount of moles of solute that appear dissolved in one liter of the mixture. In other words, molarity is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by the following expression:

$Molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$

In this case:

Molarity: 1.56 M= 1.56 $\frac{moles}{liter}$
Number of moles of calcium chlorine= ?
Volume= 2.657 liters

Replacing:

$1.56 M=\frac{Number of moles of calcium chlorine}{2.657 liters}$

Solving:

Number of moles of calcium chlorine= 1.56 M* 2.657 liters

Number of moles of calcium chlorine= 4.14 moles

In other side, you know:

Ca: 40 g/mole
Cl: 35.45 g/mole

Then the molar mass of the calcium chloride CaCl₂ is:

CaCl₂= 40 g/mole + 2* 35.45 g/mole= 110.9 g/mole

Now it is possible to apply the following rule of three: if in 1 mole there is 110.9 g of CaCl₂, in 4.14 moles of the compound how much mass is there?

$mass=\frac{4.14 moles*110.9g}{1 mole}$

mass= 459.126 g

459.126 grams of calcium chloride is needed to prepare 2.657 L of a 1.56 M solution

3 0

3 years ago

The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

$\text{Mass by mass percent}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{320.86}{1250}\times 100=25.67\%$

The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

$\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%$

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

8 0

3 years ago

Calculating Density Warm Up

Artist 52 [7]

Answer:

Explanation:

33-25=8

48/8=6

6 0

4 years ago

if 45.0 ml of 1.50 M Ca(OH)2 are needed to neutralize 25.0 ml of HI of unknown concentration, what is the molarity of the HI?

ipn [44]

Answer:

M of HI = 5.4 M.

Explanation:

We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

(XMV) acid = (XMV) base.

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

(XMV) HI = (XMV) Ca(OH)₂.

For HI; X = 1, M = ??? M, V = 25.0 mL.

For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.

∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = 5.4 M.

6 0

4 years ago