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Citrus2011 [14]
3 years ago
7

Please answer this, Determine the weight of CaO produced by heating 70 g of CaCO3

Chemistry
1 answer:
mojhsa [17]3 years ago
6 0
So first you work out the ratio. The formula tells us that for every one 'amount' of Ca and Cthere will be 3x the amount of oxygen. So, the ratio is 1:1:3, (=5) therefore 70g divided by 5 = 14.

If 70g of CaCO3 is heated there will be (1x14) 14g Ca, (1x14) 14g C and (3x14) 42g of O.

Weight of CaO= 42+14, 56g

(I hope this is right! Good luck xx)
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uranmaximum [27]

Answer:

Wavelength, \lambda=3.3\times 10^{-18}\ m

Explanation:

We have,

Velocity of the object, v=5\times 10^5\ m/s

Mass of the object, m=4\times 10^{-22}\ kg

It is required to find the wavelength of the object. The relation between wavelength and velocity of the object is given by :

\lambda=\dfrac{h}{mv}

h is Planck's constant

\lambda=\dfrac{6.6\times 10^{-34}}{4\times 10^{-22}\times 5\times 10^5}\\\\\lambda=3.3\times 10^{-18}\ m

So, the wavelength of the object is 3.3\times 10^{-18}\ m.      

5 0
3 years ago
1.Draw the skeletal structures of two different molecules that are each made of 5 carbon atoms and 12 hydrogen atoms then Name t
mihalych1998 [28]

Answer:

How many of each kind of atom is in one molecule of water? 1 hydrogen atom and 2 oxygen atoms 1 hydrogen atom and 1 oxygen atom 2 hydrogen atoms and 1 oxygen atom 2 hydrogen atoms and 2 oxygen atoms

Explanation:

1.Draw the skeletal structures of two different molecules that are each made of 5 carbon atoms and 12 hydrogen atoms then Name the two molecules you drew.

7 0
3 years ago
Water is a source of ______________ for chemical reactions in cells.
S_A_V [24]

Answer:Electrolyte

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5 0
3 years ago
The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat
Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

 Half life of lipase t_{1/2} = 8 min x 60 s/min

                                       = 480 s

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}


Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

Rate of hydrolysis V_m_{o} = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 5mmol/L

Therefore, time taken will be calculated as follows.

                    t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]

Now, putting the given values into the above formula as follows.

            t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]  

             = -\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0
3 years ago
A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a pist
REY [17]

Answer:

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

Explanation:

Determination of sign of q

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After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.

If heat is absorbed by the system, then q is indicated by positive sign and if heat is released by the system, then q is indicated by negative sign.

As in the given case, heat is released by the system, so sign of q is negative, or q < 0

Determination of sign of w

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if work is done on the system, sign of w is positive.

If work is done by the system, sign of w is negative.

In the given case, work is done on the system, therefore sign of w is positive, or w > 0

Determination of sign of ΔE

Relationship between ΔE, q and w is given by first law of thermodynamics:

ΔE = q + w

In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.

So, among the given option, option c is correct.

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

3 0
3 years ago
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