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morpeh [17]
3 years ago
14

The theory of plate tectonics was, at first, rejected by most scientists. Now most geologists accept that tectonic plates exist

and that these tectonic plates can and do move. What is responsible for the change in how scientists think about plate tectonics?
Chemistry
2 answers:
netineya [11]3 years ago
8 0

Answer:

The validation of seafloor spreading in the 1950s and 60s

Alenkasestr [34]3 years ago
4 0

Answer:

The validation of seafloor spreading in the 1950s and 60s

Explanation:

The theory of seafloor spreading was supported by numerous evidence including thermal probes that showed that heat flow over the mid-ocean ridges measured up to four times those measured in general bottom sediments, which are taken as due to the presence of molten Earth material close to the ridge crest

The ridge crest also show signs unusually seismic wave velocities that are considered to be due to microfracturing and thermal expansion from upwelling magma

You might be interested in
Given the following reaction: 2K3PO4 + AL2(CO3)3 = 3K2CO3 + 2ALPO4 If I perform this reaction with 150 g of potassium phosphate
Novosadov [1.4K]

Answer : The theoretical yield of potassium carbonate is, 146.483 g

The percent yield of potassium carbonate is, 85.33 %

Solution : Given,

Mass of K_3PO_4 = 150 g

Mass of Al_2(CO_3)_3 = 90 g

Molar mass of K_3PO_4 = 212.27 g/mole

Molar mass of Al_2(CO_3)_3 = 233.99 g/mole

Molar mass of K_2CO_3 = 138.205 g/mole

First we have to calculate the moles of K_3PO_4 and Al_2(CO_3)_3

\text{ Moles of }K_3PO_4=\frac{\text{ Mass of }K_3PO_4}{\text{ Molar mass of }K_3PO_4}=\frac{150g}{212.27g/mole}=0.7066moles

\text{ Moles of }Al_2(CO_3)_3=\frac{\text{ Mass of }Al_2(CO_3)_3}{\text{ Molar mass of }Al_2(CO_3)_3}=\frac{90g}{233.99g/mole}=0.3846moles

The given balanced reaction is,

2K_3PO_4+Al_2(CO_3)_3\rightarrow 3K_2CO_3+2AlPO_4

From the given reaction, we conclude that

2 moles of K_3PO_4 react with 1 mole of Al_2(CO_3)_3

0.7066 moles of K_3PO_4 react with \frac{1}{2}\times 0.7066=0.3533 moles of Al_2(CO_3)_3

But the moles of Al_2(CO_3)_3 is, 0.3846 moles.

So, Al_2(CO_3)_3 is an excess reagent and K_3PO_4 is a limiting reagent.

Now we have to calculate the moles of K_2CO_3.

As, 2 moles of K_3PO_4 react to give 3 moles of K_2CO_3

So, 0.7066 moles of K_3PO_4 react to give \frac{3}{2}\times 0.7066=1.0599 moles of K_2CO_3

Now we have to calculate the mass of K_2CO_3.

\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3

\text{ Mass of }K_2CO_3=(1.0599moles)\times (138.205g/mole)=146.483g

The theoretical yield of potassium carbonate = 146.483 g

The experimental yield of potassium carbonate = 125 g

Now we have to calculate the % yield of potassium carbonate.

Formula for percent yield :

\% yield=\frac{\text{ Theoretical yield}}{\text{ Experimental yield}}\times 100

\% \text{ yield of }K_2CO_3=\frac{125g}{146.483g}\times 100=85.33\%

Therefore, the % yield of potassium carbonate is, 85.33%

6 0
2 years ago
In a nuclear fusion reaction, the mass of the products is _____ the mass of the reactants.
Georgia [21]

b more than is correct answer

3 0
3 years ago
3. During an experiment where 50.0 mL of a 1.0 M acid solution was mixed with 50.0 mL of a 1.0 M base solution, the temperature
Aleksandr-060686 [28]

Answer:

\large \boxed{\text{-61 kJ$\cdot$mol$^{-1}$}}  

Explanation:

Data:

                H₃O⁺ +  OH⁻ ⟶ 2H₂O

    V/mL:  50.0     50.0

c/mol·L⁻¹:   1.0        1.0

     

    ΔT = 6.5 °C  

      ρ = 1.210 g/mL

      C = 4.18 J·°C⁻¹g⁻¹

C_cal = 12.0 J·°C⁻¹

Calculations:

(a) Moles of acid

\text{Moles of acid} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}\\\\\text{Moles of base} = \text{0.0500 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{0.0500 mol}

(b) Volume of solution

V = 50.0 mL + 50.0 mL = 100.0 mL

(c) Mass of solution

\text{Mass of solution} = \text{100.0 mL} \times \dfrac{\text{1.10 g}}{\text{1 mL}} = \text{110.0 g}

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

         q₁      +           q₂         +       q₃      = 0

     nΔH      +       mCΔT      + C_calΔT = 0

0.0500ΔH + 1.10×4.18×6.5 + 12.0×6.5 = 0

0.0500ΔH +        2989       + 78.0       = 0

                             0.0500ΔH + 3067 = 0

                                          0.0500ΔH = -3067

                                                      ΔH = -3067/0.0500

                                                            = -61 000 J/mol

                                                            = -61 kJ/mol

\text{The enthalpy of reaction is $\large \boxed{\textbf{-61 kJ$\cdot$mol$^{\mathbf{-1}}$}}$}

Note: The answer can have only two significant figures because that is all you gave for the change in temperature.

7 0
2 years ago
Consider the reaction: 3Co2+(aq) + 6NO3¯(aq) + 6Na+(aq) + 2PO43¯(aq) â Co3(PO4)2(s) + 6Na+(aq) + 6NO3¯(aq) Identify the net i
irina [24]

Answer:

D. 3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

Explanation:

The complete ionic equation includes all the ions and the insoluble species. Let's consider the following complete ionic equation.

3 Co²⁺(aq) + 6 NO₃⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 NO₃⁻(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the insoluble species. The corresponding net ionic equation is:

3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

3 0
2 years ago
Which best explains why a black T-shirt gets warmer in the Sun than a white T-shirt?
topjm [15]

Answer:

D dark colors absorb the sun more cause the wants to make everything bright I hope I helped :)

7 0
2 years ago
Read 2 more answers
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