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How many grams of calcium hydride are needed to form 4.550 g of hydrogen gas

stealth61 [152]

Molar mass

CaH₂ = 42.0 g/mol

H₂ = 2.0 g/mol

Balanced chemical equation :

CaH2 + 2 H₂O = Ca(OH)₂ + 2 H₂

42.0 g CaH₂ --------------------- 4.0 g H2
? g CaH₂ ---------------------- 4.550 g H2

mass = 4.550 * 42.0 / 4

mass = 191,1 / 4

mass = 47.775 g of CaH₂

hope this helps!

4 0

3 years ago

a gas has a volume of 350 cubic centimeters at 740 mmHg. how many cubic centimeters will the gas occupy at a pressure of 900 mmH

Harrizon [31]

Answer: 287.8 cm3

Explanation:

Given that:

Initial volume of gas V1 = 350 cm3

Initial pressure of gas P1 = 740 mmHg

New volume V2 = ?

New pressure P2 = 900 mmHg

Since, pressure and volume are involved while temperature is constant, apply the formula for Boyle's law

P1V1 = P2V2

740 mmHg x 350 cm3 = 900mmHg x V2

V2 = (740 mmHg x 350 cm3) /900mmHg

V2 = 259000 mmHg cm3 / 900mmHg

V2 = 287.8 cm3

Thus, the gas will occupy 287.8 cubic centimeters at the new pressure.

3 0

3 years ago

In certain conditions, the equilibrium concentrations for components in the reaction 2NOCl(g) 2NO(g) + Cl2(g) are [NO] = 0.02 M,

Sonja [21]

Keq= (products)/ (reactants)

Keq= ( [NO]^2 x [Cl2]) / ( [NOCl]^2)

Keq= ( (0.02)^2 x (0.01) ) / (0.5)^2= 1.6 x 10-5

8 0

3 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

VladimirAG [237]

Answer:

For 1: The standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2: The equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3: The equilibrium pressure of oxygen gas is 0.0577 atm

Explanation:

For 1:

The equation used to calculate standard Gibbs free energy change of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

For the given chemical reaction:

$M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)$

The equation for the standard Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})]$

We are given:

$\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol$

Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = ?

Putting values in above equation, we get:

$10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}$

Hence, the equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3:

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=p_{O_2}^{3/2}$

The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.

Putting values in above expression, we get:

$1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm$

Hence, the equilibrium pressure of oxygen gas is 0.0577 atm

6 0

3 years ago

How much heat is absorbed by 73.0 grams of ice as it melts at 0°C? How much heat is absorbed by 73.0 grams of liquid as it vapor

kondor19780726 [428]

Answer:

24382 J

162790 J

Explanation:

At 0°C, ice requires a certain amount of heat/energy to break its certain bonds and change its state into liquid water. This amount of heat per gram of ice is known as the latent heat of fusion of ice. And this heat is already detrmined and is equal to 334 J/g. Similarly, For water to convert into vapor at 100°C, it requires a certain amount of heat/energy to break its certain bonds and change its state into gas/vapor. This amount of heat per gram of water is known as the latent heat of vaporization of water, and it is equal to 2230 J/g.

Now, for one gram ice, 334 J heat is required for melting, hence for 73g of ice, 73x334 = 24382 J of heat is required.

Now, for one gram water, 2230 J heat is required for melting, hence for 73g of water, 73x2230 = 162790 J of heat is required.

4 0

3 years ago