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3 years ago

How many moles are in 124.7 g of Ba(OH)2

Chemistry

1 answer:

Alekssandra [29.7K]3 years ago

7 0

Answer:

0.73 mol

Explanation:

No. of moles(n)= Given mass/molar mass.

Given mass=124.7g

Molar mass of Ba(OH)2= Molar mass of (Barium+2Oxygen+2Hydrogen)=137+32+2=171g

No. of moles= 124.7g/171g=0.73 mol

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4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K

swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

$Pv =nRT$

Assuming pressure as constant,

V∝ T

Now, let K is the constant.

V = KT

Let initial volume of balloon , $V_{1}$ = 4.3 liter

1000 liter = 1 meter cube

1 liter = $\frac{1}{1000} m^{3} = 10^{-3} m^{3$

4.3 liter = $4.3\times10^{-3}=4.3\times10^{-3} m^{3}$

And initial temperature of balloon, $T_{1}$ = 350 K

Let the final volume of balloon is $V_{2}$

And as given, final temperature of balloon, $T_{2}$ is 250 K

Now, $V_{1} = KT_{1}$

$4.3\times10^{-3}=K\times350\ (equation\ 1 )$

$V_{2} = KT_{2}$

$=K\times250\ (equation 2)$

Dividing equation 1 and 2,

$\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}$

K cancelled by K.

By cross multiplication:

$350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\ = \frac{1075\times10^{-3}}{350} \\ =2.87\times10^{-3}m^{3}$

Now, convert it into liter with the help of calculation done above,

$2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter$

Therefore, volume of the gas in the balloon at 250 K will be 2.87 liter.

4 0

3 years ago

Choose the atom with:

nata0808 [166]

Answer:

a )Li

b)O

c)F

Explanation:

a) Li-1s^2 2s^1

F-1s^2 2s^2 2p^5

it is easy to pull out e- from 2p orbit than 2s because 2s orbit is close to nucleus.Therefore Li have high ionisation enthalpy

b)oxygen ion is larger than Na because o have fewer proton

c)F because it requires only 1e to achieve stable noble gas configuration.Therefore to achieve stable nobke gas electonic configuration it accept 1e.

6 0

3 years ago

For the following reaction, 28.6 grams of zinc oxide are allowed to react with 9.54 grams of water . zinc oxide(s) water(l) zinc

maw [93]

Answer:

34.9 g of Zn(OH)₂ is the maximum mass that can be formed

Explanation:

Let's state the reaction:

ZnO(s) + H₂O(l) → Zn(OH)₂ (aq)

First of all, we need to determine the moles of each reactant and state the limiting:

28.6 g . 1mol /81.38 g = 0.351 moles of ZnO

9.54 g . 1mol /18 g = 0.53 moles of water

As ratio is 1:1, for 0.53 moles of water, we need 0.53 moles of ZnO, but we only have 0.351, so the limiting reactant is the ZnO.

Ratio with the product is also 1:1. From 0.351 moles of oxide we can produce 0.351 moles of hydroxide. Let's calculate the mass:

0.351 mol . 99.4 g /1mol = 34.9 g

3 0

3 years ago

What is the ground-state electron configuration of a neutral atom of cobalt?

Andre45 [30]

4d7 , 3s2 will be the valence shell electronic configuration of a neutral atom of cobalt!

8 0

3 years ago

2 A(g) + B(g) ---> 2 C(g)Rate = k [A][B]At the beginning of one trial of this reaction, [A] = 3.0 M and [B] = 1.0 M. The obse

Orlov [11]

Answer:

Lmol⁻¹s⁻¹

Explanation:

The rate law of the given reaction is:-

Rate=k[A][B]

Wherem, k is the rate constant.

Given that:-

Rate = 0.36 mol/Lsec = 0.36 M/sec

[A] = 3.0 M

[B] = 1.0 M

Thus,

Applying in the equation as:-

0.36 M/sec =k × 3.0 M× 1.0 M

k = 0.12 (Ms)⁻¹ = 0.12 Lmol⁻¹s⁻¹

<u>The units of k = Lmol⁻¹s⁻¹</u>

6 0

3 years ago