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dimulka [17.4K]
3 years ago
8

In the following reaction, which substance is the precipitate? (NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq) (NH4)2SO4 Ba(

NO3)2 BaSO4 NH4NO3
Physics
2 answers:
Arte-miy333 [17]3 years ago
5 0

Answer:

BaSo4

Explanation:

Mariana [72]3 years ago
4 0
I think it's Barium sulfate, the soild and percipitate
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I have been trying to do number 2 and I just don’t know what the answer is if you can help me thank you so mush
skelet666 [1.2K]

Answer:

Strong Positive Linear Line Correlation

Explanation:

According to the diagram, there is a strong positive linear line correlation between velocity and distance of galaxies. Hubble showed that galaxies are receding away from people with a velocity that is proportional to their distance from people more distant galaxies recede faster than nearby galaxies.

About the diagram: Velocity is the distance relation among extra-galactic nebulae. Radial velocities is corrected for solar motion, are plotted against distances estimated from involved stars and mean luminosities of nebulae in a cluster. The black discs and full line represent the solution for solar motion by using the nebulae individually; the circles and broken line represent the solution combining the nebulae into groups; the cross represents the mean velocity corresponding to the mean distance of 22 nebulae whose distances could not be estimated individually.

4 0
2 years ago
It's a gun made of skin and the bullet is made of gas it is natural ​
NNADVOKAT [17]

Answer:

Is IT A Ruger

Explanation:

wHaT

4 0
4 years ago
What is the mass of an 291pound persond
Luden [163]
Well, it also depends on the height... but say if they were 5'3" and 291 pounds...
their BMI would be

51.55
4 0
3 years ago
Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
salantis [7]

Explanation:

Mass of bumper cars, m_1=m_2=110\ kg

Initial speed of car A, u_1=8\ m/s

Initial speed of car Z, u_2=-10\ m/s

Final speed of car A after the collision, v_1=-10\ m/s

We need to find the velocity of car Z after the collision. Let it is equal to v_2. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

110\times 8+110\times (-10)=110\times (-10)+110v_2

v_2=\dfrac{-1320}{110}\ m/s

v_2=-12\ m/s

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

5 0
3 years ago
A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of the E
Sveta_85 [38]

Answer:

995 N

Explanation:

Weight of surface, w= 4000N

Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

m= 4000/9.81= 407.7472

Using radius of orbit of 6371km

The force of gravity of satellite in its orbit, F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}

Where G=6.67*10^{-11} and M=5.94*10^{24}

F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}

F= 995.01142 then rounded off

F=995N

6 0
3 years ago
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