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2 years ago

Which directly contribute to sea level rise?(1 point) It's actually earth science.

Chemistry

2 answers:

Rudiy272 years ago

8 0

Answer:?Any one have the answers?

Explanation:d

faust18 [17]2 years ago

3 0

Answer:

1. reflected energy and melting sea ice*

2. The institutions gathered enough data for scientists to draw a conclusion about whether the annual global temperature is gradually increasing.

The data gathered by each institution is very similar to the data gathered by the other institutions.

Explanation:

The reflection of extra heat causes the sea ice to melt. This causes the sea ice to melt. This becomes a problem because the polar ice melts and contributes to the increase in the water content in the sea. This causes the rise in the sea levels.

The options (1) and (2) are correct because the institutions gathered enough data to reach plausible conclusions.

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What process would farmers use to produce vegetables that will grow in a specific climate?

netineya [11]

They would make sure it gets lots of water and it needs sunlight to so some kind of source that could replace the sun

4 0

3 years ago

Hydrogenated fats are produced by passing hydrogen gas over long saturated carbon chains. A

REY [17]

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8 0

3 years ago

Calculate the heat needed to increase the temperature of 100. g water from 45.7 C to 103.5 C.

MA_775_DIABLO [31]

Answer:

Total heat required to raise the temperature of water from 45.7°C to 103.5°C

= 249,362.4 J

Explanation:

The Heat required to raise the temperature of 100.0 g of water from 45.7°C to 103.5°C will be a sum of;

- The heat required to raise the 100 g of water from 45.7°C to water's boiling point of 100°C

- The Heat required to vaporize the 100 g of water at its boiling point

- The Heat required to raise the temperature of this vapour from 100°C to 103.5°C

1) The heat required to raise the 100 g of water from 45.7°C to water's boiling point of 100°C

Q = mCΔT

m = 100 g

C = 4.18 J/g.°C

ΔT = change in temperature = (100 - 45.7) = 54.3°C

Q = 100 × 4.18 × 54.3 = 22,697.4 J

2) The Heat required to vaporize the 100 g of water at its boiling point

Q = mL

m = 100 g

L = ΔHvaporization = 2260 J/g

Q = mL = 100 × 2260 = 226,000 J

3) The Heat required to raise the temperature of this vapour from 100°C to 103.5°C

Q = mCΔT

m = 100 g

C = 1.90 J/g.°C

ΔT = change in temperature = (103.5 - 100) = 3.5°C

Q = 100 × 1.9 × 3.5 = 665 J

Total heat required to raise the temperature of water from 45.7°C to 103.5°C

= 22,697.4 + 226,000 + 665

= 249,362.4 J

Hope this Helps!!!

4 0

3 years ago

Read 2 more answers

Un tecnico di laboratorio deve preparare una soluzione di carbonato di sodio decaidrato, Na2CO3⋅ 10 H2O per eseguire alcune anal

fiasKO [112]

Answer:

1. 3.70 g Na₂CO₃·10H₂O

2. 50.0 mL of the first solution

Explanation:

1. Prepare the solution

(a) Calculate the molar mass of Na₂CO₃·10H₂O

$\begin{array}{rrr}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 2\times22.99 & 45.98\\\text{1C} & 1\times 12.01 & 12.01\\\text{13O}&13 \times16.00 & 208.00\\\text{20H}&20 \times 1.008 & 20.16\\&\text{TOTAL =} & \mathbf{286.15}\\\end{array}$

The molar mass of Na₂CO₃·10H₂O is 286.15 g/mol.

(b) Calculate the moles of Na₂CO₃·10H₂O

$\text{Moles of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\= \text{0.250 L solution} \times \dfrac{\text{0.0500 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 L solution}}\\\\= \text{0.0125 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

(c) Calculate the mass of Na₂CO₃·10H₂O

$\text{Mass of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O }\\= \text{0.012 50 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O } \times \dfrac{\text{296.15 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}{\text{1 mol Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}}\\\\= \text{3.70 g Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}\\\text{You need $\large \boxed{\textbf{3.70 g}}$ of Na$_{2}$CO$_{3}\cdot$10H$_{2}$O}$

2. Dilute the solution

We can use the dilution formula to calculate the volume needed.

V₁c₁ = V₂c₂

Data:

V₁ = ?; c₁ = 0.0500 mol·L⁻¹

V₂ = 100 mL; c₂ = 0.0250 mol·L⁻¹

Calculation:

$\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\V_{1}\times \text{0.0500 mol/L} & = & \text{100 mL} \times\text{0.0250 mol/L}\\0.0500V_{1}& = & \text{2.500 mL}\\V_{1}&=& \dfrac{\text{2.500 mL}}{0.0500}\\\\& = & \text{50.0 mL}\\\end{array}\\\text{You need $\large \boxed{\textbf{50.0 mL}}$ of the concentrated solution.}$

7 0

3 years ago

What is hydrogen bonding

yKpoI14uk [10]

Answer:

a weak bond between two molecules resulting from an electrostatic attraction between a proton in one molecule and an electronegative atom in the other.

Explanation:

For example, in water molecules (H2O), hydrogen is covalently bonded to the more electronegative oxygen atom. Therefore, hydrogen bonding arises in water molecules due to the dipole-dipole interactions between the hydrogen atom of one water molecule and the oxygen atom of another H2O molecule.

4 0

3 years ago