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tigry1 [53]
3 years ago
7

M = 50 kg

Chemistry
1 answer:
Hatshy [7]3 years ago
5 0

Answer:

<h3>The answer is option A</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

where

f is the force

m is the mass

a is the acceleration

From the question

f = 150 N

m = 50 kg

We have

a =  \frac{150}{50}  =  \frac{15}{5}  \\

We have the final answer as

<h3>3 m/s²</h3>

Hope this helps you

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Answer:

We assume you are converting between moles CaCl2 and gram. You can view more details on each measurement unit: molecular weight of CaCl2 or grams This compound is also known as Calcium Chloride. The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles CaCl2, or 110.984 grams.

7 0
3 years ago
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Mechanical twinning occurs in metals having which type(s) of crystal structure(s)?
kari74 [83]
Mechanical twinning occurs in metals having body center cubic and hexagonal closed packed structures. Twinning is said to occur when a portion of a crystal takes up an orientation that is related to the orientation of the untwinned lattice in a definite symmetrical manner. 
5 0
3 years ago
When you are converting grams to moles , are you multiplying or dividing ?
Effectus [21]
When you are converting grams to moles, the operation that you will be doing is dividing by the molar mass to obtain the amount of moles of the given sample.
4 0
3 years ago
If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio
bulgar [2K]

Answer:

m_{Cd(OH)_2}=36.6 gCd(OH)_2

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2}  =0.75molCd(OH)_2

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2

Best regards.

4 0
3 years ago
A sealed 1.0L flask is filled with 0.500 mols of I_2 and 0.500 mols of Br_2. When the container achieves equilibrium the equilib
DochEvi [55]

Answer:

[IBr] = 0.049 M.

Explanation:

Hello there!

In this case, according to the balanced chemical reaction:

I_2+Br_2\rightarrow 2IBr

It is possible to set up the following equilibrium expression:

K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110

Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of x (reaction extent) would be:

0.0110=\frac{(2x)^2}{(0.50-x)^2}

Which can be solved for x to obtain two possible results:

x_1=-0.0277M\\\\x_2=0.0245M

Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:

[IBr]=2x=2*0.0249M=0.049M

Regards!

7 0
2 years ago
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