All categories

2 years ago

How many moles are in 0.50 g sodium bromide

1 answer:

6 0

Use stoichometry to calculate this. In order to calculate this answer, you need to divide the grams by the molar mass of sodium bromide. The molar mass of sodium bromide is approximately 103 g/mol. 0.50/103 equates to about 0.00485 moles of sodium bromide.

You might be interested in

Calculate the molarity of H3PO4 when you added 57.3 g into 3,820 mL of water

motikmotik

Answer:

0.153M

Explanation:

57.3/97.994 (molar mass)=0.585 moles of H3PO4

.0585/3.820L=0.153M

5 0

4 years ago

2ch4(g) c2h2(g) 3h2(g) describe what is happening within the system when it is at equilibrium in terms of concentrations, reacti

SashulF [63]

Balanced chemical reaction: 2CH₄(g) ⇄ C₂H₂(g) + 3H₂(g).

1) In a chemical reaction, chemical equilibrium is the state in which both reactants (methane CH₄) and products (ethyne C₂H₂ and hydrogen H₂) are present in concentrations which have no further tendency to change with time.

2) At equilibrium, both the forward and reverse reactions are still occurring.

3) Reaction rates of the forward and backward reactions are equal and there are no changes in the concentrations of the reactants and products.

4 0

3 years ago

Read 2 more answers

A student carries out a flame test on an unknown solid. a red flame is seen. the student concludes that the solid is lithium car

Sergio [31]

I think this is the answer you was looking for hope this helps !!!

3 0

3 years ago

Pulverized coal pellets, which may be approximated as carbon spheres of radius ro= 1 mm, are burned in a pure oxygen atmosphere

Ganezh [65]

Answer:

Explanation:

SO; If we assume that:

P should be the diffusion of oxygen towards the surface ; &

Q should be the diffusion of carbondioxide away from the surface.

Then the total molar flux of oxygen is illustrated by :

$Na,x = - cD_{PQ}\frac{dy_P}{dr} +y_P(NP,x + N_Q,x)$

where;

r is the radial distance from the center of the carbon particle.

Since ;

$N_P,x = - N_Q, x$ ; we have:

$Na,x = - cD_{PQ}\frac{dy_P}{dr}$

The system is not steady state and the molar flux is not independent of r because the area of mass transfer $4\pi r^{2}$ is not a constant term.

Therefore, using quasi steady state assumption, the mass transfer rate $4\pi r^{2}N_{P,x}$ is assumed to be independent of r at any instant of time.

$W_{P}=4\pi r^{2}N_{P,x}$

$W_{P}=-4\pi r^{2}cD_{PQ}\frac{dy_{P}}{dr}$

= constant

The oxygen concentration at the surface of the coal particle $yP,R$ will be calculated from the reaction at the surface.

The mole fraction of oxygen at a location far from pellet is 1.

Thus, separating the variables and integrating result into the following:

$W_{P}\int_{R}^{\infty} \frac{dr}{r^{2}}=-4\pi$

$r^{2}cD_{PQ}\int_{y_{P,R}}^{y_{P,\infty }}dy_{P}$

$-W_{P}\frac{1}{r}\mid ^{\infty }_{R}= -4\pi cD_{PQ}(y_{P,\infty }-y_{P,R})$

$=> W_{P}= - 4\pi cD_{PQ}(1-y_{P,R})R$

The mole of oxygen arrived at the carbon surface is equal to the mole of oxygen consumed by the chemical reaction.

$W_{P} = 4 \pi R^2R"$

$W_{P}= 4\pi R^{2}k_{1}"C_{O_{2}}\mid _{R}$

$W_{P}= 4\pi R^{2}k_{1}"c y _{P,R}$

$-4\pi cD_{PQ}(1-y_{P,R})R= - 4\pi R^{2}k_{1}"c y _{P,R}$

$y_{P,R}=\frac{D_{PQ}}{D_{PQ}+Rk_{1}}$

$y_{P,R}=\frac{1.7 \times 10^{-4}}{1.7\times 10^{-4}+10^{-3}\times 0.1}$

$\mathbf{= 0.631}$

Obtaining the total gas concentration from the ideal gas law; we have the following:

where;

R= $0.082m^3atm/kmolK$

$c=\frac{P}{RT} \\ \\ c=\frac{1}{0.082\times 1450} \\ \\ = 0.008405kmol/m^3$

The steady state $O_2$ molar consumption rate is:

$W_{P}= -4\pi cD_{PQ}(1-y_{P,R})R$

$W_{P}= -4\pi (0.008405)(1.7\times 10^{-4})(1-0.631)(10^{-3})$

$W_{P}= - 6.66\times 10^{-9}kmol/s$

5 0

3 years ago

Question 18 of 32

guajiro [1.7K]

Answer: A. A bond in which molecules share electrons

8 0

3 years ago