Answer: 1.85M
Explanation:
Sodium hydroxide has a chemical formula of NaOH
Given that,
Amount of moles of NaOH (n) = ?
Mass of NaOH in grams = 63g
For molar mass of NaOH, use the molar masses:
Sodium, Na = 23g;
Oxygen, O = 16g;
Hydrogen, H = 1g
NaOH = (23g + 16g + 1g)
= 40g/mol
Since, amount of moles = mass in grams / molar mass
n = 63g / 40.0g/mol
n = 1.575 mole
Volume of NaOH solution (v) = 850mL
[Convert 850mL to liters
If 1000mL = 1L
850mL = 850/1000 = 0.85L]
Concentration of NaOH solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 1.575 mole / 0.85L
c = 1.85M [1.85M is the concentration in moles per litres which is also known as molarity]
Thus, the molarity of the solution is 1.85M
10, deca means ten a decagon is a ten side polygon, a decimetre is one tenth of a metre a decade is ten years so your answer is ten.
Hola!
→ Your query:
Atoms of one element easily changes to atoms of another element. Is it True? or False?
→ The answer is: TRUE
→ Reason :
We can change atom of one element to other atom by adding more particles into them. ex : Neutron, Protons and Electrons.
This is how nuclear fusion works.
hope it helps!
Answer: 5.26 moles
Explanation:
Given that:
Volume of hydrogen gas V = 8560mL
(since 1000 mL = 1dm3
8560mL = 8560/1000= 8.56dm3)
Standard temperature T = 25°C
Convert temperature in Celsius to Kelvin
(25°C + 273 = 298K)
Pressure P = 1.5atm
Number of moles of hydrogen = ?
Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
1.5 atm x 8.56dm3 = n x (0.0082 atm dm3 K-1 mol-1 x 298K)
12.84atm dm3 = n x 2.44atm dm3 mol-1
n = (12.84atm dm3 / 2.44atm dm3 mol-1)
n = 5.26 moles
Thus, there are 5.26 moles of Hydrogen, H present in this gas sample.

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it used to convey information from one place to another through intervening media,like- Radios and televisions

used in hospitals to produce photograph of bone to check for break or fracture.They can penetrate less dense matter such body tissue and skin.