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Maru [420]
3 years ago
7

In MOST cases, a hurricane will

Chemistry
2 answers:
Nady [450]3 years ago
7 0

Answer:

Increase its energy when coming in contact with warm water

Explanation:

kari74 [83]3 years ago
6 0

Answer:increase it energhy when coming in contact with warm water

Explanation:because they dont form in cold places

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What is the mass of 21.7 moles of Al(NO3)3?
Elza [17]

Answer:I would say 21grams but i am not sure

Explanation:

5 0
3 years ago
Calculate the molality of a solution formed by adding 6.30 g NH4CL to 15.7 g of water
babymother [125]

Answer:

Molality = 7.5 mol/kg

Explanation:

Given data:

Mass of NH₄Cl = 6.30 g

Mass of water = 15.7 g (15.7/1000 =0.016 kg)

Molality = ?

Solution:

Formula of molality:

Molality = Moles of solute / mass of solvent in gram

Now we will first calculate the number of moles of solute( NH₄Cl )

Number of moles = mass/ molar mass

Molar mass of  NH₄Cl = 53.491 g/mol

Number of moles = 6.30 g/  53.491 g/mol

Number of moles =  0.12 mol

Now we will calculate the molality.

Molality = Moles of solute / mass of solvent in gram

Molality =  0.12 mol / 0.016 kg

Molality = 7.5 m

or        (m=mol/kg)

Molality = 7.5 mol/kg

6 0
3 years ago
Find the formula for the hydrate<br>0.737 g MgSO3 and 0.763 g H2O
babunello [35]

The required formula of hydrate is MgSO₃.6H₂O.

<h3>How do we calculate the formula of hydrate?</h3>

The number of moles of water per mole of anhydrous solid (x) will be computed by dividing the number of moles of water by the number of moles of anhydrous solid (x) to find the hydrate's formula.

Moles will be calculated as:
n = W/M, where

  • W = given mass
  • M = molar mass

Moles of MgSO₃ = 0.737g / 104.3g/mol = 0.007mol

Moles of H₂O = 0.763g / 18g/mol = 0.04 mol

Number of H₂O molecule = 0.04/0.007 = 5.7 = 6

So formula of hydrate is MgSO₃.6H₂O.

Hence required formula of hydrate compound is MgSO₃.6H₂O.

To know more about hydrate compound, visit the below link:

brainly.com/question/22411417

#SPJ1

6 0
2 years ago
The conversion of cyclopropane to propene occurs with a first-order rate constant of . How long will it take for the concentrati
Gennadij [26K]

This is an incomplete question, here is a complete question.

The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

Answer : The time taken will be, 17.0 hr

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.42\times 10^{-2}\text{ hr}^{-1}

t = time passed by the sample  = ?

a = initial concentration of the reactant  = 0.080 M

a - x = concentration left = 0.053 M

Now put all the given values in above equation, we get

t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}

t=17.0\text{ hr}

Therefore, the time taken will be, 17.0 hr

5 0
3 years ago
Express the answer in scientific notation and with the correct number of significant figures: (6.32 x 10-4) ÷ 12.64
elixir [45]
<span>Express the answer in scientific notation and with the correct number of significant figures:
(6.32 x 10-4) ÷ 12.64
5.00 x 10^-5</span>
4 0
4 years ago
Read 2 more answers
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