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Fofino [41]
3 years ago
5

Find the mass of 1.112 moles HF?

Chemistry
1 answer:
Alexeev081 [22]3 years ago
7 0

Answer:

22.25 g

Explanation:

To find the mass, you need to convert moles to grams and get 22.25 g.

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Will ya'll help me, please
JulsSmile [24]

Answer:

Either A or D

But i think D

hope this helps =3

sorry if im wrong

7 0
3 years ago
What are the units of k in the following rate law?
Ludmilka [50]

Answer:

B. \frac{1}{M^{2} s }

Explanation:

The unit for rate is M/s while the unit for each molecule should be M. You can find the unit for k by putting the units for rate and the molecules into the equation

rate= k{X][Y]

M/s= k * M^{2} * M^{1}

k= (M/s) / (M^{3})

k= \frac{1}{M^{2} s }

You can also use this predetermined formula to solve this problem faster: k= \frac{M^{1-n} }{s }

Where n is the number of molecule. There are 3 molecule(2X and 1Y) so n=3, so

k= \frac{M^{1-n} }{s }

k= \frac{M^{1-3} }{s }= \frac{m^{-2}}{s}= \frac{1}{M^{2} s }

8 0
4 years ago
At ground level, how are rain and hail similar?
Varvara68 [4.7K]

Answer:

rain is water in molten state

hails are water in solid form.

hope it helps

Plz mark me as brainleist if helps

Have a great day ahead

3 0
3 years ago
Read 2 more answers
A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain
tatiyna

Answer:

C_m=0.474\frac{J}{g\°C}

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

Q_m=-Q_w

Thus, in terms of masses, specific heats and temperatures we can write:

m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}

Best regards!

6 0
3 years ago
Which statement is true of a rock layer that had another rock layer deposited on top of it?​
Karo-lina-s [1.5K]

Answer:

second rock...I know it

6 0
3 years ago
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