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2 years ago

¿Cuántos moles hay en 350 g de NaNO3?

Chemistry

1 answer:

olga55 [171]2 years ago

4 0

Answer:

3 moles

Explanation:

So, approximately 3 moles of NaNO3 can be obtained, by reacting 253 grams of Na2CrO4. Also, the number of moles is a dimensionless quantity.

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What percent of Earths feshwater is found as a solid?

Zanzabum

Answer:

68.7% is the percent of frozen freshwater.

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2 years ago

What sphere is most diffrent from earth

goblinko [34]

Answer:

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3 years ago

Lithium bromide dissociates in water according to the following thermochemical equation: LiBr(s) → Li+ (aq) + Br– (aq) ΔH = –48.

madreJ [45]

Answer:

Final temperature of water is $48.3^{0}\textrm{C}$

Explanation:

1 mol of LiBr releases 48.83 kJ of heat upon dissolution in water.

So, 2 moles of LiBr release $(2\times 48.83)kJ$ or 97.66 kJ of heat upon dissolution in water.

This amount of heat is consumed by 1000.0 g of water. Hence temperature of water will increase.

Let's say final temperature of water is $t^{0}\textrm{C}$ .

So, change in temperature ( $\Delta T$ ) of water is $(t-25)^{0}\textrm{C}$ or (t-25) K

Heat capacity (C) of water is $4.184\frac{J}{g.K}$

Hence, $m_{water}\times C_{water}\times \Delta T_{water}=97.66\times 10^{3}J$

where m is mass

So, $(1000.0g)\times (4.184\frac{J}{g.K})\times (t-25)K=97.66\times 10^{3}J$

or, $t=48.3$

Hence final temperature of water is $48.3^{0}\textrm{C}$

3 0

3 years ago

For the decomposition of ammonia on a platinum surface at 856 °C

Crazy boy [7]

$\\ \tt\leadsto \dfrac{d[NH_3]}{dt}=1.50\times 10^{-6}$

dt remains same for reaction

$\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}\dfrac{d[NH_3]}{dt}$

$\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}(1.5\times 10^{-6})$

$\\ \tt\leadsto \dfrac{d[H_2]}{dt}=2.25\times 10^{-6}Ms^{-1}$

M is molarity here not metre

5 0

1 year ago

Ammonia (NH3) gas and oxygen (O2) gas react to form nitrogen (N2) gas and water (H2O) vapor. Suppose you have 2.0 mol of NH3 and

IrinaK [193]

Answer: 3.5 mol of oxygen

Explanation:

The unbalanced equation for this reaction is

$\text{NH}_{3}+\text{O}_{2} \longrightarrow \text{N}_{2}+\text{H}_{2}\text{O}$

Balancing this equation,

$4\text{NH}_{3}+3\text{O}_{2} \longrightarrow 2\text{N}_{2}+6\text{H}_{2}\text{O}$

From this, we can tell that for every 4 moles of ammonia consumed, 3 moles of oxygen are consumed.

Considering the ammonia, the reaction can occur 2.0/4 = 1/2 a time.
Considering the oxygen, the reaction can occur 3/5 = 3/5 a time.

This means that ammonia is the limiting reactant, meaning that 2.0(3/4)=1.5 moles of oxygen are consumed.

So, 5.0-1.5=<u>3.5 mol</u> of oxygen remain.

6 0

1 year ago