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3 years ago

13

Are protons unstable

1 answer:

kiruha [24]3 years ago

5 0

If you're including protons that are bound as part of an atomic nucleus, then, yes, under certain conditions.

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1-<br> the<br> Nalt and Gil-<br> Cu2+<br> ad out- <br> what are the spectator ions in this reaction

Mrac [35]

Answer:

cu2 is 5he correct answer

3 0

1 year ago

Can someone please help

hram777 [196]

Answer: usuhduhduhdjhd

Explanation:

jaajaajaj

8 0

2 years ago

What substance is reduced in the reaction 2Fe 2+ + CI2 = 2Fe 3+ + 2CI- ?

atroni [7]

Answer:

$Cl_{2}$ is reduced in the reaction

Explanation:

The given reaction is

$2Fe^{2+}+Cl_{2} \to 2Fe^{3+}+2Cl^{-}$

The oxidation number of $Fe$ is changed from $+2\, \to \, +3$

$Fe^{2+} \to Fe^{3+}+e^{-}$

And The oxidation number of $Cl$ is changed from $0\, \to \, -1$

$Cl_{2}^{0} + 2e^{-} \to 2Cl^{-}$

Hence, $Fe^{2+}$ is oxidized and $Cl_{2}$ is reduced

5 0

3 years ago

The chemical formula of ordinary sugar is C12H22O11. Calculate the mass of 7.35 mol of sugar.

ch4aika [34]

First, you need to find the mass of 1 mol of sugar. Mass, or molar mass, can simply be found by adding the masses of the individual elements. These are given to you on the periodic table.

$C_{12}H_{22}O_{11}$

12 x 12.011 grams (molar mass of Carbon) = 144.132 g

22 x 1.008 grams (molar mass of Hydrogen) = 22.176 g

11 x 15.999 grams (molar mass of Oxygen) = 175.989 g

Add all of the pieces together.

144.132 g + 22.176 g + 175.989 g = 342.297 grams

So, if one mole has 342.297 grams, then 7.35 of that amount will be your answer.

342.297 g/mol x 7.35 mol = 2,515.88 grams

7 0

3 years ago

n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains

Nitella [24]

Answer : The metal used was iron (the specific heat capacity is $0.449J/g^oC$ ).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of unknown metal = ?

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of unknown metal = 150 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature of water = $34.3^oC$

$T_1$ = initial temperature of unknown metal = $150.0^oC$

$T_2$ = initial temperature of water = $25.0^oC$

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

$150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC$

$c_1=0.449J/g^oC$

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is $0.449J/g^oC$ ).

6 0

3 years ago