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baherus [9]
3 years ago
5

How much does calcite coast

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

Answer:

It depends on what kind and how much.  Some are about $100, and others are $1,000.

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Do liquids have higher coefficients of expansion than solids do
zmey [24]

Answer: yes they do.

Explanation:

I'm tryna get more points because my other account got blocked for the next 48 hours.

(Don't be mad at me please)

4 0
3 years ago
A gas exerts a pressure of 0.62 atm. Convert this to kPa and mmHg. Be sure to show your work.
IRINA_888 [86]

Answer:

The answer to your question is 0.62 atm = 62.82 kPa = 471.2 mmHg

Explanation:

Data

P = 0.62 atm

P = ? kPa

P = ? mmHg

Process

1.- Look for the conversion factor of atm to kPa and mmHg

 1 atm = 101.325 kPa

1 atm = 760 mmHg

2.- Do the conversions

                  1 atm ----------------- 101.325 kPa

                  0.62 atm ------------  x

                   x = (0,62 x 101.325) / 1

                  x = 62.82 kPa

                   1 atm ------------------ 760 mmHg

                   0.62 atm ------------  x

                   x = (0.62 x 760)/1

                   x = 471.2 mmHg                

4 0
3 years ago
How many formula units make up 19.0 g of magnesium chloride (MgCl2)?
professor190 [17]
To convert the mass of a compound to formula units, the conversion factor is Avogadro's number, 6.022 x10^23 formula units/ mol and its molar mass. In this case, we are given with 19.0 grams of magnesium chloride which has a mass of 95.21 g/mol. Hence the answer is 1.20 x 10^23 formula units.
5 0
3 years ago
Heyy pls help anyone?
madam [21]
1a. calcium chloride (CaCl2)
b. 2HCl (aq) + Ca(OH)2 (s) —> CaCl2 (aq) + 2H2O (l)

i’m not sure about the rest but i hope this helped ^^
7 0
3 years ago
Read 2 more answers
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
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