A sample of metal had a mass of 17.29g and a volume of 5.68mL what is the density of this metal

What Is Ethane? Can you help me?

arsen [322]

Ethane is gas used as fuel: a colorless odorless gas that is highly flammable.

I hope this answered your question

8 0

3 years ago

Read 2 more answers

Describe how wave frequency changes across the electromagnetic spectrum, from radio waves to gamma rays

qwelly [4]

I believe the way it works is the frequency increases as it goes from radio waves to gamma rays, because the energy is increasing.

7 0

2 years ago

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0

2 years ago

Ammonium phosphate ((NH4)3 PO4) is an important ingredient to many fertilizers. It can be made by reacting phosphoric acid (H3 P

lara31 [8.8K]

Answer:

15.35 g of (NH₄)₃PO₄

Explanation:

First we need to look at the chemical reaction:

3 NH₃ + H₃PO₄ → (NH₄)₃PO₄

Now we calculate the number of moles of ammonia (NH₃):

number of moles = mass / molecular wight

number of moles = 5.24 / 17 = 0.308 moles of NH₃

Now from the chemical reaction we devise the following reasoning:

if 3 moles of NH₃ are produce 1 mole of (NH₄)₃PO₄

then 0.308 moles of NH₃ are produce X moles of (NH₄)₃PO₄

X = (0.308 × 1) / 3 = 0.103 moles of (NH₄)₃PO₄

mass = number of moles × molecular wight

mass = 0.103 × 149 = 15.35 g of (NH₄)₃PO₄

4 0

2 years ago

Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have

Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

7 0

2 years ago