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AlexFokin [52]
3 years ago
11

Two common ions of manganese are Mn2+ and Mn3+. Which binary ionic compound is likely to form? A) Mn2N3 B) MnBr4 C) MnCl D) MnO

Physics
2 answers:
Varvara68 [4.7K]3 years ago
5 0

Answer: D

Explanation: The answer is D

meriva3 years ago
4 0
Given that Oxygen has an oxidation state of  2 -, you can combine Mn 3+ with O 2- ions to form Mn2O3, and you can combine Mn 2+ with O 2- to form Mn2O2 which is MnO.

The other compounds imply oxidation states of N, Br and Cl that does not exist.

Therefore, the answer is the option D. MnO
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Estimate the kinetic energy (in GJ) of a 93,000 metric ton aircraft carrier moving at a speed of at 32 knots. You will need to l
Wewaii [24]

Answer:

kinetic Energy = 12.58 GJ

Explanation:

1 metric ton is equal to 1000 kg

then,

93000 metric ton

Mass is m = 93000 x 1000 kg

speed is v = 32 knots

1 knot is 0.514 m/s

then ,

v = 32 x 0.514 = 16.448 m/s

To solve for the Kinetic Energy (KE), we have;

KE = 0.5 x m x v²

KE = 0.5*93000*1000*(16.448)²

     = 12.58 x 10^{9} J

      = 12.58 GJ

6 0
3 years ago
What does area under a velocity time graph represent
Feliz [49]
Velocity hope that helps
7 0
3 years ago
Ninas measurements shown in the table here BEST represent a wave with
Finger [1]
They best represent a wave with zero energy and zero amplitude.

There are no measurements shown in a table that accompanies
this question having any amplitude or energy greater than zero.
3 0
3 years ago
Read 2 more answers
How to calculate percent error
olga_2 [115]

Answer:

Steps to Calculate the Percent Error

Subtract the accepted value from the experimental value.

Take the absolute value of step 1.

Divide that answer by the accepted value.

Multiply that answer by 100 and add the % symbol to express the answer as a percentage.

Explanation:

3 0
3 years ago
The air in a room with volume 180 m3 contains 0.25% carbon dioxide initially. fresher air with only 0.05% carbon dioxide flows i
Butoxors [25]

Answer:

p(t)=\frac{1}{20}(1+2e^{-t/90})

Explanation:

Let p(t) be % of CO_2 at time t(time in minutes).

Amount of CO_2 in room=Volume of room \times p

Rate of change of CO_2 in room=Volume of room x \frac{dp}{dt}

=180\times \frac{dp}{dt}

Rate of inflow of CO_2=(\% CO_2 \ in\  Fresh\  Air \times \frac{1}{100})\times(Rate \ of  Inflow\ of \ air)

=(0.05\times \frac{1}{100})\times 2=\frac{1}{100}

Rate of CO_2 outflow

(\% CO_2 \ in\  Fresh\  Air \times \frac{1}{100})\times(Rate \ of  Outflow\ of \ air)\\=(p\times\frac{1}{100})\times2=\frac{p}{50}

Therefore rate of change of CO_2 is \frac{1}{100}-\frac{p}{50}

% rate of change is100\times \ Rate of \ Change=\frac{1}{10}-2p

Therefore we have:

180\frac{dp}{dt}=\frac{1}{10}-2p\\\frac{dp}{dt}=\frac{1-20p}{1800}\\\\\frac{dp}{20p-1}=-\frac{dt}{1800}

Integrate both:

\int\limits \frac{dp}{20p-1}=\int\limits-\frac{dt}{1800}

\frac{1}{20}In|20p-1|=-\frac{t}{1800}+In \ C

In|20p-1|=-\frac{t}{90}+InK

+20In \ C=+In \ K

Raise both sides to base e,

20p-1=e^{-\frac{t}{90}+In \ K}\\\\p=\frac{1}{20}(1+Ke^{-t/90})

Initially, there's only 0.25% of CO_2,

Now substitute p=0.25 and t=0

0.25=\frac{1}{20}(1+K)\\K=4

p=\frac{1}{20}(1+2e^{-t/90})

8 0
3 years ago
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