All categories

3 years ago

Find the molar mass of Ca(NO3)2

Chemistry

2 answers:

spayn [35]3 years ago

6 0

The molar mass of Ca(NO₃)2 is 164 g/mol

Explanation

Ca(NO₃)2 is made up of Ca , N and O elements

Therefore the molar mass of Ca(NO₃)2 = molar mass of Ca + molar mass of N +molar mass of O

from periodic table the molar mass of Ca is 40 g/mol

for N = 14 g/ mol

for O = 16 g/mol. since they are 3 atoms of O in Ca(NO₃)2 the molar mass = 16 x3 = 48 g/mol

The molar mass of Ca(NO₃)2 = 40 g/mo + [(14 g/mol + 48 g/mol) 2] =164g/mol

sdas [7]3 years ago

6 0

Answer:164.10 to be exact

Explanation:

You might be interested in

Explain how chemical change differs from a physical change

Grace [21]

A chemical change affects on the molecular level of matter, which makes it irreversible. Combustion is a pretty good exmple. Physical changes are reversible and dont alter the formula. Hope this helped!

8 0

3 years ago

Read 2 more answers

How many valence electrons should you use to draw a Lewis structure of ammonia (NH3)?

Mekhanik [1.2K]

Hello there,

Your correct answer would be 8. There are 8 valence electrons available for the Lewis structure for NH3.

Hope this helps

~Hottwizzlers

4 0

3 years ago

Read 2 more answers

True [87]

P1V1 = P2V2

P1 = 720 mmHg
V1 = 450. mL
P2 = 760 mmHg (this is the pressure at STP)

Use these to solve for V2:
(720)(450) = 760V2

V2 = 426 mL

3 0

2 years ago

Read 2 more answers

When the forward and reverse paths of a change occur at the same rate,

Zarrin [17]

The correct answer is option B. When the forward and reverse paths of a change occur at the same rate, the system is in equilibrium specifically in dynamic equilibrium. Dynamic equilibrium is the balance in a process that is continuing. It is achieved in a reaction when the forward rate of reaction and the backward rate of reaction is at the same value or equal.

8 0

3 years ago

Read 2 more answers

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0

3 years ago