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Vanyuwa [196]
4 years ago
12

Which of these peices off evidence did Alfred Wegners original theory of continental drift have access to?

Chemistry
2 answers:
alekssr [168]4 years ago
3 0
Hi Alex,
I think your answer is (<span>a seafloor that was geologically active with earthquakes, volcanoes, and mountain chains) Good luck!
:)</span>
lisabon 2012 [21]4 years ago
3 0

The answer is A seafloor that was geologically active with earthquakes, volcanoes, and mountain chains.

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The molar mass of carbon dioxide (CO2) is 44.01 g/mol. The molar mass of water (H2O) is 18.01 g/mol. A reaction uses 528 g of CO
Nataly [62]
The reaction that results from this is:

H2O + CO2 --> H2CO3

Stoichiometric ratio between water and CO2 is 1:1. So we can say that for every Mole of CO2, we need 1 Mole of water to produce 1 Mole of H2CO3. Thus as n=m/M we can find n = 528/44.01 = 11.997 ~ 12Mol.

therefore, we need 12 moles of water.
5 0
4 years ago
Read 2 more answers
In an ionic compound, the charges of the _ and _ must balance to produce an electrically substance.
solong [7]

Answer:

positive(cations) and negative(anions)

Explanation:

5 0
3 years ago
How many valence electrons are in the methyl ammonium ion CH3NH3+
Anika [276]

Answer:

15

Explanation:

8 0
3 years ago
Does O2 have a single, double, or triple bond?
miss Akunina [59]
Answer: has a double bond
Explanation: O2 has a double bond in its normal form. That is O=O. There are no unpaired electrons in this case are there since there are 2 lone pairs on each oxygen.
4 0
3 years ago
A is a solution containing 3.5g of HX per dm-³. B is a solution containing 0.050moldm-³ of an hydrous sodium trioxocarbonate (IV
Cerrena [4.2K]

The Molar Concentration Of A =0.099 .

The Relative Molar Mass Of A = 35.0129 gm

Given,

Mass of HX = 3.5 g

Moles of solution B (Na_{2} CO_{3}) = 0.05 moles

Volume of HX = 26.10 mL

Volume of Solution B = 25 mL

Molecular weight of solution B = 2(atomic weight of Na )+ atomic weight of C + 3(atomic weight of O)

                 = 2(23) + 12 + 3(16)

                 =106 gm

Equivalent weight of  Na_{2} CO_{3} = molecular weight / 2 = 106 /2  =53 g

Mole = mass / molecular weight

∴0.05 = mass / 106

∴ mass = 5.3 gm Na_{2} CO_{3}

Normality = mass ÷ (equivalent weight × volume of the solution in liter)

                = 5.3 ÷( 53 × 0.025)

                =4 N

So, by using formula ,

N_{1} V_{1} =N_{2} V_{2}

N_{1} = normality of solution B = 4 N

V_{1} = volume of solution B = 25 mL

N_{2} = normality of solution A = ? N

V_{2} =Volume of solution A = 26.1 mL

∴ 4×25 = N_{2} × 26.1

∴N_{2} = 3.83 N

∴ normality of solution A = 3.83 N

from Formula of the normality we can find the equivalent weight of the A

Normality = mass of HX ÷ (equivalent weight × volume of the solution in liter)

3.83 = 3.5 ÷( equivalent weight × 0.0261)

∴equivalent weight = 35.0129 g

In case of HX the electron transfer is 1 ,so equivalent weight = molecular weight ; which is also termed as relative molar mass in given case.

∴The Relative Molar Mass Of A = 35.0129 g

Molar concentration = mass / molar mass

                               = 3.5 / 35.0129

                               = 0.099 mole

∴ The Molar Concentration of A  is 0.099 .

Learn more about Normality here...

brainly.com/question/1685497

#SPJ10

3 0
2 years ago
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