IUPAC name for BaSO3

An aqueous solution of sodium thiosulfate releases heat when it crystallizes. What would happen when sodium thiosulfate crystals

rusak2 [61]

Answer:

An excellent experiment is to heat sodium thiosulfate in a water bath. The solid crystals will dissolve into the water in the hydrated crystals forming a supersaturated solution. ... Placing a small crystal in the supersaturated solution will cause the liquid to turn solid.

7 0

3 years ago

When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x

Leno4ka [110]

You have to use the equation PV=nRT.
P=pressure (in this case 1.89x10^3 kPa which equals 18.35677 atm)
1V=volume (in this case 685L)
n=moles (in this case the unknown)
R=gas constant (0.08206 (L atm)/(mol K))
T=temperature (in this case 621 K)
with the given information you can rewrite the ideal gas law equation as n=PV/RT.
n=(18.35677atm x 685L)/(0.08206atmL/molK x 621K)
n=246.8 moles

8 0

3 years ago

If 5,800 j of energy are applied to a 15.2 kg piece of lead, by how much does the temp change if the specific heat of lead is 0.

scoray [572]

Answer:

The answer will be 2.98K

Explanation:

Using the formula:

Q = mc∆T

Q= 5,800 (heat in joules)

m= convert 15.2kg to g which is 15200g (mass in grams)

c= 0.128 J/g °c (Specific heat capacity)

∆T= what we need to find (temperature change)

5800J = 15200g x 0.128 x ∆T

= 2.98K

7 0

4 years ago

In the laboratory you dissolve 13.9 g of potassium phosphatein a volumetric flask and add water to a total volume of 250mL.

tensa zangetsu [6.8K]

Answer:

Molarity of the solution? 0,262 M.

Concentration of the potassium cation? 0,786 M.

Concentration of the phosphate anion? 0,262 M.

Explanation:

Potassium phosphate (K₃PO₄; 212,27 g/mol) dissolves in water thus:

K₃PO₄ → 3 K⁺ + PO₄³⁻ <em>(1)</em>

Molarity is an unit of chemical concentration given in moles of solute (K₃PO₄) per liters of solution.

There are 250 mL of solution≡0,25 L

The moles of K₃PO₄ are:

13,9 g of K₃PO₄ × $\frac{1mol}{212,27 g}$ = 0,0655 moles of K₃PO₄

The molarity of the solution is:

$\frac{0,0655 moles}{0,25L}$ = 0,262 M

In (1) you can see that 1 mole of K₃PO₄ produces 3 moles of potassium cation. The moles of potassium cation are:

0,0655 moles×3 = 0,1965 moles

The concentration is:

$\frac{0,1965 moles}{0,25L}$ = 0,786 M

The moles of K₃PO₄ are the same than moles of PO₄³⁻, thus, concentration of phosphate anion is the same than concentration of K₃PO₄. 0,262 M

I hope it helps!

4 0

3 years ago

A solution of which of the following coordination compounds will form a precipitate when treated with aqueous AgNO3?

lidiya [134]

Answer:

[Cr(NH3)6.]C13

Explanation:

Alfred Werner's coordination theory (1893) recognized two kinds of valency;

Primary valency which are nondirectional and secondary valency which are directional.

Hence, the number of counter ions precipitated from a complex depends on the primary valency of the central metal ion in the complex.

We must note that it is only these counter ions that occur outside the coordination sphere that can be precipitated by AgNO3.

If we consider the options carefully, only [Cr(NH3)6.]C13 possess counter ions outside the coordination sphere which can be precipitated when treated with aqueous AgNO3.

3 0

3 years ago