3 years ago

Which scientist was first to extensively develop the theory of continental drift?

Chemistry

2 answers:

fiasKO [112]3 years ago

7 0

Answer:

Alfred Wegener

Explanation:

K12

Simora [160]3 years ago

4 0

A. Alfred Wegener

hope that helps you :) ✨

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How many milliliters of 12.0 m hcl(aq) are needed to prepare 905.0 ml of 1.00 m hcl(aq)?

givi [52]

The ml of 12.0 M HCL needed to prepare 905.0 ml of 1.00m hcl is calculated using M1V1 =M2V2 formula
M1= 12.0 M
V1=?
M2= 905.0 ml
V2 = 1.00M

V1 is therefore = M2V2/M1

=905 x 1.00/12.0 = 75.42 ml

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3 years ago

What is the molarity of a solution that contains 11.6 moles of LiNO3 in 13.7 liters of solution?

Tatiana [17]

The molarity of this should be around 0.846

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Formula: M= mols/liters

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So… M= 11.6/13.7⇒ 0.846

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2 years ago

the solubility product Ag3PO4 is: Ksp = 2.8 x 10^-18. What is the solubility of Ag3PO4 in water, in moles per liter?

guapka [62]

Answer : The solubility of $Ag_3PO_4$ in water is, $1.8\times 10^{-5}mol/L$

Explanation :

The solubility equilibrium reaction will be:

$Ag_3PO_4\rightleftharpoons 3Ag^++PO_4^{3-}$

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ag^{+}]^3[PO_4^{3-}]$

$K_{sp}=(3s)^3\times (s)$

$K_{sp}=27s^4$

Given:

$K_{sp}$ = $2.8\times 10^{-18}$

Now put all the given values in the above expression, we get:

$K_{sp}=27s^4$

$2.8\times 10^{-18}=27s^4$

$s=1.8\times 10^{-5}M=1.8\times 10^{-5}mol/L$

Therefore, the solubility of $Ag_3PO_4$ in water is, $1.8\times 10^{-5}mol/L$

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2 years ago

I need to know what’s the error in each equation help !!!

lorasvet [3.4K]

Good luck my guy your going to need it

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2 years ago

If hydrogen-1 has 1 proton how many protons does hydrogen-2 have?

Vikentia [17]

Number of protons remain the same however number of neutron increases. <span />

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3 years ago