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dem82 [27]
2 years ago
11

What is the name of the following compound? Se5S6

Chemistry
1 answer:
inessss [21]2 years ago
5 0

Answer:

I'm sorry I couldn't find it, I looked all over the internet

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What poisonous gas is found in the exhaust fumes of car engines?
Ugo [173]
The toxic gar expelled from the reaction between gasoline and oxygen in the vehicle's engine is both Carbon dioxide and monoxide
4 0
3 years ago
Read 2 more answers
The volume of a sample of nitrogen is 6.00 liters at 35oC and 0.75 atm. What volume will it occupy at STP?
joja [24]
The  volume that  will  occupy  at STP  is  calculated   as  follows
by use  of ideal  gas  equation
that is  PV=nRT  where n  is  number of  moles  calculate  number of moles

n= PV/RT
p=0.75 atm
V=6.0  L
R = 0.0821  L.atm/k.mol
T=  35  +273= 308k
n=?

n=  (o.75  atm  x  6.0 L)/( 0.0821 L.atm/k.mol  x 308 k)=  0.178 moles

Agt  STP  1 mole=  22.4 L  what obout  0.178 moles

=  22.4  x0.178moles/ 1moles =3.98 L( answer C)
3 0
3 years ago
BRAINLIESTTT ASAP!!! PLEASE HELP ME :)
SCORPION-xisa [38]

Every science experiment should follow the basic principles of proper investigation so that the results presented at the end are seen as credible.

Observation and Hypothesis. ...

Prediction and Modeling. ...

Testing and Error Estimation. ...

Result Gathering and Presentation. ...

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4 0
3 years ago
Two objects are brought into contact Object 1 has mass 0.76 kg, specific heat capacity 0.87) g'c and initial temperature 52.2 'C
taurus [48]

Answer:

T_F=77.4\°C

Explanation:

Hello there!

In this case, according to the given information, it turns out possible to set up the following energy equation for both objects 1 and 2:

Q_1=-Q_2

In terms of mass, specific heat and temperature change is:

m_1C_1(T_F-T_1)=-m_2C_2(T_F-T_2)

Now, solve for the final temperature, as follows:

T_F=\frac{m_1C_1T_1+m_2C_2T_2}{m_1C_1+m_2C_2}

Then, plug in the masses, specific heat and temperatures to obtain:

T_F=\frac{760g*0.87\frac{J}{g\°C} *52.2\°C+70.7g*3.071\frac{J}{g\°C}*154\°C}{760g*0.87\frac{J}{g\°C} +70.7g*3.071\frac{J}{g\°C}} \\\\T_F=77.4\°C

Yet, the values do not seem to have been given correctly in the problem, so it'll be convenient for you to recheck them.

Regards!

4 0
3 years ago
How much heat is needed warm 65.34 g of water from 18.43 to 21.75
BaLLatris [955]

Given that  

Mass of water = 65.34 g

Amount of heat = mass of water * specific heat (temperature change

)

= 65.34 g * 4.184 J / g-C ( 21.75-18.43 )C

= 907.63  J  

= 0.908 KJ

And  

1 cal = 4.186798 J

907.63 J * 1 cal / 4.186798 J =216.78  cal

Or0.218 kcal  


5 0
3 years ago
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