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Vilka [71]
2 years ago
10

A student Increases the temperature of a 417 cubic centimeter balloon from 278 K to 231 K. Assuming constant pressure what shoul

d the new volume of the balloon be
Chemistry
1 answer:
WITCHER [35]2 years ago
5 0
Initial Conditions:
Volume= v1= 417 cm³
Temperature= T1 = 278 K
 
Final Conditions:
Temperature= T2 = 231K
Volume = v2 =?

Use the general gas equation;

                                        P1*v1/T1 =  P2*v2/T2
As, the temperature is constant;
 So,
                                               v1/T1 = v2/T2
                                              417/278 = v2/231
                                                  v2= 346.5 cm³                                           
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If 200.0 g of copper(II) sulfate react with an excess of zinc metal, what is the theoretical yield of copper? 1.253 g 50.72 g 79
Helen [10]
1)we need a balanced equation: CuSO₄ + Zn ---> ZnSO₄ + Cu

2) we need to convert the grams of CuSO₄ to moles using the molar mass. 

molar mass CuSO₄= 63.5 + 32.0 + (4 x 16.0)= 160 g/mol

200.0 g CuSO_4 ( \frac{1 mol}{160 grams} )= 1.25 mol CuSO_4

3) convert moles of CuSO₄ to moles of Cu

1.25 mol CuSO_4 ( \frac{1 mol Cu}{1 mol CuSO_4} )= 1.25 mol Cu

4) convert moles of Cu to grams using it's molar mass.

molar mass Cu= 63.5 g/mol

1.25 mol (\frac{63.5 grams}{1 mol} )= 79.4 grams Cu

I did it step-by-step as the explanation but you can do all of this in one step. 

200.0 g CuSO_4 ( \frac{1 mol CuSO_4}{160 g} ) ( \frac{1 mol Cu}{1 mol CuSO_4} ) ( \frac{63.5 grams}{1 mol Cu} )= 79.4 grams Cu


4 0
2 years ago
Select the NMR spectrum that corresponds best to p-bromoaniline. (see Hint for the structure.) The selected tab will be highligh
lana [24]

Answer:

The NMR spectrum that corresponds best to p-bromoaniline  is the one that is attached in the image below.

Explanation:

For the p-bromoaniline 3 types of hydrogen are observed. The first signal that appears at 3.7 ppm would be from the hydrogens of the NH2 group, the hydrogens in ortho position with respect to the NH2 group give a double at approximately 6.54 ppm, and finally the characteristic 7.21 ppm signal is observed for the hydrogens in meta position with with respect to the NH2 group.

7 0
2 years ago
How many molecules of glucose are in 1 l of a 100 mm glucose solution?
Karolina [17]

Answer is: 6.022·10²² molecules of glucose.

c(glucose) = 100 mM.

c(glucose) = 100 · 10⁻³ mol/L.

c(glucose) = 0.1 mol/L; concentration of glucose solution.

V(glucose) = 1 L; volume of glucose solution.

n(glucose) = c(glucose) · V(glucose).

n(glucose) = 0.1 mol/L · 1 L.

n(glucose) = 0.1 mol; amount of substance.

N(glucose) = n(glucose) · Na (Avogadro constant).

N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.

N(glucose) = 6.022·10²².

6 0
2 years ago
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