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PtichkaEL [24]
3 years ago
8

Rewrite this measurement with a simpler unit, if possible. 8.3 kg•m^2/ s^2

Chemistry
1 answer:
Feliz [49]3 years ago
7 0

Answer: 8.3 J

Explanation:

We have the following measurement:

8.3 \frac{kg m^{2}}{s^{2}}

Rearranging the units:

8.3 \frac{kg m}{s^{2}}m

Since 1 Newton is 1 N=1 \frac{kg m}{s^{2}}:

8.3 Nm

Since 1 Joule is 1 J=1 Nm:

8.3 J This is the simplest form possible

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Write the formulas for the following ionic compounds: (a) copper bromide (containing the Cu+ ion), (b) manganese oxide (containi
ivann1987 [24]

Answer:The formulas of ionic compounds are:

a)CuBr

b)Mn_2O_3

c)Hg_2I_2

d)Mg_3(PO_4)_2

Explanation:

Formulas for the an ionic compounds is determine by:

Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

(a) Copper bromide :Given that it contains Cu^+ ion.

Cu^++Br^-\rightarrow CuBr

(b) Manganese oxide : Given that it contains Mn^{3+} ion.

Mn^{3+}+O^{2-}\rightarrow Mn_2O_3

(c)Mercury iodide :Given that it contains Hg_2^{2+}

Hg_2^{2+}+I^-\rightarrow Hg_2I_2

(d) Magnesium phosphate :Given that it contains PO_4^{3-}

Mg^{2+}+PO_4^{3-}\rightarrow Mg_3(PO_4)_2

4 0
2 years ago
33 grams of NH4NO3 will produce how many grams of water
Leni [432]

Answer:

66 grams

Explanation:

For every 1 gram of NH4NO3 equals 2 grams of water

8 0
3 years ago
What is the epicenter?
hoa [83]

Answer:

basically the middle of an earthquake

5 0
3 years ago
In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
How did your estimated poputation size<br> compare with the actual population size
musickatia [10]

Answer:

Hope this helps

Explanation:

https://seagrant.whoi.edu/wp-content/uploads/2018/05/ESTIMATING-POPULATION-SIZE-1.pdf

8 0
2 years ago
Read 2 more answers
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