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e-lub [12.9K]
4 years ago
10

In any pure sample of CO2, the ratio of the mass of carbon to the mass of

Chemistry
1 answer:
Vlad [161]4 years ago
7 0

Answer:A

Explanation:por q si bb

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The ionic mobility of alkali metal ions in aqueous<br>solution is maximum for​
matrenka [14]

Answer:

Rb+

hope it's right:)

..................................

8 0
3 years ago
Calculate the percentage composition of Mg3 (Po4)2
lutik1710 [3]
Mg3(PO4)2 - the molar mass would be 262g/mol, which is 100%

Atomic mass of Mg is 24, since we have 3Mg we multiply by 3 and get a mass of 72

262 : 100% = 72 : x%

x = 72*100 / 262

x = 27.5%

And do that for every element — get the molar mass of P and multiply by 2, use a ratio, and get the molar mass of O and multiply by 8 and use ratios :)
7 0
3 years ago
A 100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution before the addition of any lioh
snow_lady [41]
Thank you for posting your question here at brainly. Below is the solution:

 <span>moles HClO4 = 0.100 L x 0.18 M = 0.018 
moles LiOH = 0.030 L x 0.27 = 0.0081 
moles H+ in excess = 0.018 - 0.0081 = 0.0099 
total volume = 0.130 L 
[H+] = 0.0099/ 0.130= 0.0762 M 
pH = 1.12</span>
3 0
3 years ago
How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
xenn [34]

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

Best regards!

4 0
3 years ago
Convert 59,800 cg/L to g/mL
alexgriva [62]
To convert the given value to the desired one, use the proper unit conversions and dimensional analysis. Use the following conversion for the first set.

    1 g = 100 cg
    1 L = 1000 mL

Using the concept presented above,
         
            V = (59800 cg/L)(1 g/100 cg)1 L/1000 mL)
            V = 0.598 g/mL
6 0
3 years ago
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