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Marina86 [1]
2 years ago
6

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

Chemistry
1 answer:
Charra [1.4K]2 years ago
3 0

Answer:

heres the link of the answer

boldniwally.com

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Calculate the grams of CaCl2 necessary to make a 0.15Msolution.
xxTIMURxx [149]

Answer: mass m = M·c·V

Explanation: M(CaCl2) = 110.98 g/mol, c= 0.15 mol/l,

n=m/M= cV, volume of Solution is not mentioned

3 0
3 years ago
A balloon occupies a volume of 2.00 l at 40.0oc. how much volume will it occupy at 30.0oc?
hodyreva [135]
<span>Charles' law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".

V </span>α T

Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

V₁ = 2.00 L
T₁ = 40.0 ⁰C = 313 K
V₂ = ?
T₂ = 30.0 ⁰C = 303 K

By applying the formula,
2.00 L / 313 K = V₂ / 303 K                   
                   V₂ = (2.00 L / 313 K) x 303 K
                   V₂ = 1.94 L

Hence, the volume of the balloon at 30.0 ⁰C  is 1.94 L
7 0
3 years ago
What is the mass of an empty film canister?
Alekssandra [29.7K]
If this is just a general question it seems to vary from about 4.5g to 5g. Is there more data to the question?
6 0
3 years ago
How much heat, in calories, is given off when 1.25 grams of silver is cooled from 100.0 degrees Celsius to 80 degrees Celsius? (
lapo4ka [179]
The heat required to raise the temperature of a certain mass of sample to a specific temperature change, we use the formula mCpΔT where m is mass, Cp is the specific heat of the substance and ΔT is the temperature change. In this case, we substitute and form 1.25 g x 0.057 cal/g C *20 C equal to 1.425 calories.
7 0
3 years ago
Determine the empirical formula of a compound containing 1.71 g of silicon and 8.63 g of chlorine.
Basile [38]

Answer:

The answer to your question is: SiCl₄

Explanation:

Data

amount of Si      1.71 g

amount of Cl     8.63 g

MW Si = 28 g

MW Cl = 35.5

Process (rule of three)

For Si                                                        For Cl

        28 g of Si ------------------ 1 mol                      35.5 g of Cl --------------- 1 mol

          1.71g of Si  ---------------   x                              8.63 g of Cl --------------  x

         x = 1.71 x 1 / 28 = 0.06 mol                          x = 8.63 x 1 / 35.5 = 0.24 mol

Now, divide both results by the lowest of them.

Si = 0.06 mol / 0.06 = 1 molecule of Si     Cl = 0.24 / 0.06 = 4 molecules of Cl

Finally

                     Si₁ Cl₄ or SiCl₄

8 0
3 years ago
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