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3 years ago

How many moles of H2 are needed to produce 34.8 moles of NH3?

Chemistry

1 answer:

lesya692 [45]3 years ago

7 0

2 i hope this helps

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A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree

lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = $T_1=33.9^oC=33.9+273K=306.9 K$

Final temperature of the water = $T_2$

Specific heat capacity of water under these conditions = c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

$Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)$

$-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC$

The new temperature of the water bath 32.0°C.

5 0

3 years ago

2. Name each of the following ionic compounds A. K2O B. Cacl2. C. Mg3N2 D. NaCIO E. KNO3

Ludmilka [50]

A. Potassium oxide
B. Calcium chloride
C. Magnesium nitride
D. Sodium hypochlorite
E. Potassium nitrate

3 0

3 years ago

What substance can be used to electrolyze water? any electrolyte that is not easily reduced or oxidized hydrochloric acid only s

Alisiya [41]

<span>any electrolyte that is not easily reduced or oxidized</span>

7 0

3 years ago

What is the molarity of the solution produced when 8.0g NaOH is dissolved in sufficient water to prepare 500.mL of solution?

Sindrei [870]

Answer:

Molarity = 0.4M

Explanation:

Molar mass of NaOH (M)= 40

m= 8g, V= 500ml=0.5L

n= m/M=[8/40]= 0.2mol

Applying

n= CV

0.2= C×0.5

C= 0.4M

8 0

3 years ago

A 0.15 m solution of chloroacetic acid has a ph of 1.86. What is the value of ka for this acid?

dem82 [27]

Answer: $1.67\times 10^{-3}$

Explanation:

$ClCH_2COOH\rightarrow ClCH_2COO^-+H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given: c = 0.15 M

pH = 1.86

$K_a$ = ?

Putting in the values we get:

Also $pH=-log[H^+]$

$1.86=-log[H^+]$

$[H^+]=0.01$

$[H^+]=c\times \alpha$

$0.01=0.15\times \alpha$

$\alpha=0.06$

As $[H^+]=[ClCH_2COO^-]=0.01$

$K_a=\frac{(0.01)^2}{(0.15-0.15\times 0.06)}$

$K_a=1.67\times 10^{-3]$

Thus the vale of $K_a$ for the acid is $1.67\times 10^{-3}$

4 0

3 years ago