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DanielleElmas [232]
3 years ago
10

What are the 3 forces that make the tectonic plates move

Chemistry
1 answer:
Molodets [167]3 years ago
8 0

Answer:

Viscous Drag.

Slab-Pull Force.

Ridge-Push Force

Explanation:

your welcome have a nice day

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Assume that 12.0 g of oxygen are reacted with 20.0 g of magnesium to produce magnesium oxide.
FinnZ [79.3K]
Write out the eqn of magnesium and oxygen. this should be under “metals” chapter. do revise.

next, find the mols of both oxygen and magnesium. compare the ratios and find the LIMITING REAGENT.

use the mols of the limiting reagent to compare with the mols of the product.

take the mols of the product/mr of the product.

this will give u the mass.
8 0
3 years ago
Which describes the composition of carbohydrates?
UkoKoshka [18]

Answer: Carbohydrates (carbo- = “carbon”; hydrate = “water”) contain the elements carbon, hydrogen, and oxygen, and only those elements with a few exceptions. The ratio of carbon to hydrogen to oxygen in carbohydrate molecules is 1:2:1.

HOPE THIS HELPS

CAN U GIVE ME BRAINLIEST

3 0
2 years ago
Does 1 gram of phosphorus react with 6 grams of iodine to form 4 grams of phosphorus triodine in P4(s)+6I2(s)=4PI3(s)
mafiozo [28]

Answer:

No

Explanation:

One mole of P₄ react with six moles of I₂ and gives 4 moles of PI₃.

When one gram phosphorus and 6 gram of  iodine react they gives 8.234 g ram of PI₃ .

Given data:

Mass of phosphorus = 1 g

Mass of iodine = 6 g

Mass of  PI₃ = ?

Solution:

Chemical equation:

P₄ + 6I₂    →  4PI₃

Number of moles of P₄:

Number of moles = Mass /molar mass

Number of mole = 1 g / 123.9 g/mol

Number of moles  = 0.01 mol

Number of moles of I₂:

Number of moles  = Mass /molar mass

Number of moles = 6 g / 253.8 g/mol

Number of moles = 0.024 mol

Now we will compare the moles of PI₃ with I₂ and P₄.

                I₂              :              PI₃

                  6              :               4

                 0.024       :             4/6×0.024 = 0.02

                  P₄            :               PI₃

                 1                :                4

                 0.01          :               4 × 0.01 = 0.04  mol

The number of moles of PI₃ produced by I₂ are less it will be limiting reactant.

Mass of PI₃ = moles × molar mass

Mass of PI₃ = 0.02 mol × 411.7 g/mol

Mass of PI₃ =  8.234 g

4 0
3 years ago
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

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

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
3 years ago
A very loud sound has a high _____.
DENIUS [597]

Answer:

Resting Point

Explanation:

4 0
2 years ago
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