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Mkey [24]
2 years ago
15

Calculate the number of atoms in a 3.62 × 10³ g sample of strontium. ​

Chemistry
1 answer:
oksian1 [2.3K]2 years ago
5 0

hope it helps you ...........

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25. 0 ml of a 0. 100 m solution of NH3 is titrated with 0. 250m HCl. After 10. 0 ml of the hcl has been added, the resultant sol
AveGali [126]

Main Answer: The resultant solution will have 0.125M of HCl.

Explanation:

Given:

Volume of NH3  V1= 25ml = 0.025lit

Molar concentration of NH3  M1= 0.1M

It is titrated with HCl

Molar concentration of HCl M2= 0.25M

Volume of HCl  V2= ?ml

At equilibrium,

M1V1 = M2V2

V2 = M1V1/M2

V2 = (0.1*0.025)/0.25

V2= 0.01lit =0.01*1000 = 10ml

Number of moles = M2*V2 = 0.25 * 0.01 = 0.0025

Now 10ml of HCl is added.

then,

V2 = 10 + 10 = 20ml

Molar concentration of HCl = number of moles/Volume

                                               = 0.0025/0.02

                                               = 0.125M

The resultant solution will have 0.125M of HCl.

To know more about resultant solution, please visit:

brainly.com/question/20367152

#SPJ4

5 0
1 year ago
What are 5 steps of balancing a chemical equation plz help me
nlexa [21]

Answer:

Explanation:

Identify each element found in the equation. The number of atoms of each type of atom must be the same on each side of the equation once it has been balanced.

What is the net charge on each side of the equation? The net charge must be the same on each side of the equation once it has been balanced.

If possible, start with an element found in one compound on each side of the equation. Change the coefficients (the numbers in front of the compound or molecule) so that the number of atoms of the element is the same on each side of the equation. Remember, to balance an equation, you change the coefficients, not the subscripts in the formulas.

Once you have balanced one element, do the same thing with another element. Proceed until all elements have been balanced. It's easiest to leave elements found in pure form for last.

Check your work to make certain the charge on both sides of the equation is also balanced.

6 0
3 years ago
A tank at is filled with of dinitrogen difluoride gas and of chlorine pentafluoride gas. You can assume both gases behave as ide
jenyasd209 [6]

The question is incomplete, the complete question is;

A 8.00 L tank at 2.64 °C is filled with 9.82 g of chlorine pentafluoride gas and 10.1 g of dinitrogen difluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Be sure your answers have the correct number of significant digits.

Answer:

See explanation for details

Explanation:

Number of moles of N2F2 = mass/ molar mass

Molar mass of N2F2 = 66 g/mol

Number of moles = 10.1 g/66 = 0.15 moles

Number of moles of ClF5 = 9.82 g/130 g/mol= 0.08 moles

Molar mass of ClF5= 130g/mol

Total number of moles = 0.15 moles + 0.08 moles = 0.23 moles

Given that;

T= 2.64 °C + 273 = 275.64 K

n= 0.23 moles

R= 0.082 Latmmol-1K-1

V= 8.00 L

P= ??

From;

PV =nRT

P= nRT/V

P= 0.23 ×0.082 × 275.64/8.00

P= 0.65 atm

Mole fraction of N2F2= 0.15/0.23 = 0.65

Partial pressure = mole fraction × total pressure = 0.65 × 0.65 = 0.42 atm

Mole fraction of ClF5 = 0.08/0.23 = 0.35

Partial pressure of ClF5 = mole fraction × total pressure = 0.35 × 0.65 = 0.22 atm

6 0
3 years ago
What is the percentage of water in the hydrate cocl2 • 6h2o
lord [1]

Answer:

52.17%

Explanation:

COCl2.6H20

C=12,O=16,Cl=35.5,H=1

Relative molecular mass of COCl2.6H2O= 12+16+71+6(2+16) = 99 + 108= 207g

Relative molecular mass of 6H2O = 108g

Percentage of water = (108/207 )*100

= 52.17%

8 0
3 years ago
What is the with of the tank in cm that is0.20 meters
Marat540 [252]
So, the answer would be 20 cm

8 0
3 years ago
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