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4vir4ik [10]
3 years ago
13

Please help!! my grades dying 15 points

Chemistry
1 answer:
AysviL [449]3 years ago
4 0

Answers-in-bold:

There are two common temperature scales. On the Fahrenheit scale, water freezes at 32 degrees. The Celsius scale divides the interval between the freezing and boiling points of water into 100 degrees.

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Which phrase describes air density?
saw5 [17]

Answer:B equals mass divided by volume

Explanation:I got hacks :)

7 0
3 years ago
Read 2 more answers
CAN SOME ONE HELLLLLLLLLLLLLLLLLLLLLLP WILL GIVE BRAINLIEST TO THE BEST ANSWR FOR THIS AND GIVE AWAY 15 POINTS
lord [1]
1. C
2. C
3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit. 
4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust.  Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress. 

I hope this helped! These are COMPLEX questions though! =D
8 0
3 years ago
If a sample containing 18.1 g of NH3 is reacted with 90.4 g of
USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃  = 18.1g

Mass of Cu₂O  = 90.4g

Unknown:

Limiting reactant  = ?

Mass of N₂ formed  = ?

Solution:

The reaction equation is given as:

       Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

  Number of moles = \frac{mass}{molar mass}  

Molar mass of Cu₂O = 2(63.6) + 16  = 143.2g/mol

Molar mass of NH₃  = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = \frac{18.1}{143.2}   = 0.13moles

Number of moles of NH₃   = \frac{90.4}{17}   = 5.32moles

  From this reaction;

       1 mole of  Cu₂O combines with 2 mole of NH₃

So   0.13moles of  Cu₂O will combine with 0.13 x 2 mole of NH₃

                                              = 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

   Mass = number of moles x molar mass

    1 mole of Cu₂O  will produce 1 mole of N₂

    0.13 mole of Cu₂O  will produce 0.13 mole of N₂

    Mass  = 0.13 x (2 x 14) = 3.64g

5 0
3 years ago
For the reaction shown, calculate how many grams of each product form when the following amounts of reactant completely react to
stira [4]
This may help you
<span>You need to use some stoichiometry here. The only way to do that is if you're working in moles. Since you're given grams of Al, you can convert that moles by dividing by the molar mass. Then from looking at the coefficients in your equation, you can see that for however many moles of Al react, the same numbers of moles of Fe will be produced, but only half as many moles of Al2O3 will be produced. To go back to grams, multiply the moles of each product that you get by their molar masses!</span>
7 0
3 years ago
calculate q for the following: 125.0 ml of 0.0500 m pb(no3)2 is mixed with 75.0 ml of 0.0200 m nacl at 25oc chegg
alexandr402 [8]

The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.

Given, T = 25°C.

<h3>Chemical equation:</h3>

Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2

PbCl2 in aqueous solution split into following ions

PbCl2 ------ Pb(+2) + 2Cl-

Q = [Pb(+2)] [Cl-]^2

The Concentration of Pb(+2) ions and Cl- ions can be calculated as

[Pb(+2)] = 0.06 × 125/200

= 0.0375

[Cl-] = 0.02 × 75/200

= 0.0075

By substituting all the values, we get

[0.0375] [0.0075]^2

= 2.11 × 10^(-6).

Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).

learn more about Ions:

brainly.com/question/13692734

#SPJ4

6 0
2 years ago
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