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3 years ago

How many moles are in 579 grams of Pt

Chemistry

2 answers:

aleksandr82 [10.1K]3 years ago

6 0

2.96804 moles that’s the answer. hope it helped

Tasya [4]3 years ago

5 0

Answer:

Use the formula n=m/M and you’ll find that there are 2.967 moles in 579 g of platinum .

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The partial pressure of cartbon dioxide in the atmgsphere is 0239 toer Caiculate the partial pressure in mm He and atm Rpund eac

Verizon [17]

Answer :

The pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.

The pressure of carbon dioxide in the atmosphere in atm is $3.14\times 10^{-4}atm$ .

Explanation :

The conversion used for pressure from torr to mmHg is:

1 torr = 1 mmHg

The conversion used pressure from torr to atm is:

1 atm = 760 torr

or,

$1torr=\frac{1}{760}atm$

As we are given the pressure of carbon dioxide in the atmosphere 0.239 torr. Now we have to determine the pressure of carbon dioxide in the atmosphere in mmHg and atm.

<u>Pressure in mmHg :</u>

As, $1torr=1mmHg$

So, $0.239torr=\frac{0.239torr}{1torr}\times 1mmHg=0.239mmHg$

Thus, the pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.

<u>Pressure in atm:</u>

As, $1torr=\frac{1}{760}atm$

So, $0.239torr=\frac{0.239torr}{1torr}\times \frac{1}{760}atm=3.14\times 10^{-4}atm$

Thus, the pressure of carbon dioxide in the atmosphere in atm is $3.14\times 10^{-4}atm$ .

3 0

3 years ago

Balancing the reaction by oxidation number method k2cr2o7+sncl2+hcl

Mumz [18]

Answer:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

Explanation:

The products of this reaction are given by:

$K_2Cr_2O_7 (aq) + SnCl_2 (aq) + HCl (aq)\rightarrow KCl (aq) + SnCl_4 (aq) + CrCl_3 (aq) + H_2O (l)$

Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

$Cr_2O_7^{2-} (aq)\rightarrow Cr^{3+} (aq)$

The following steps should be taken:

balance the main element, chromium: multiply the right side by 2 to get 2 chromium species on both side:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)$

balance oxygen atoms by adding 7 water molecules on the right:

$Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the hydrogen atoms by adding 14 protons on the left:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

Similarly, tin(II) cation becomes tin(IV) cation:

$Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-$

Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)$

$3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-$

Add them together:

$Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 3 Sn^{2+} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l) + 3 Sn^{4+} (aq)$

Adding the ions spectators:

$K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)$

7 0

3 years ago

How do forces in nuclear and chemical reactions differ

marusya05 [52]

1) Nuclear reactions involve a change in an atom's nucleus, usually producing a different element. Chemical reactions, on the other hand, involve only a rearrangement of electrons and do not involve changes in the nuclei. ... (3) Rates of chemical reactions are influenced by temperature and catalysts.

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3 years ago

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Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

Anastasy [175]

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

7 0

3 years ago

If the ph of a solution is decreased from ph 8 to ph 6 it means that the

saw5 [17]

Explanation:

Relation between pH and concentration of hydrogen ions is as follows.

pH = $-log [H^{+}]$

So, it means that an increase in the value of pH will show that there occurs a decrease in concentration of hydrogen ions.

Therefore, the solution becomes basic in nature.

On the other hand, a decrease in the value of pH will show that there occurs an increase in the concentration of hydrogen ions.

Therefore, the solution becomes more acidic in nature.

Hence, if the pH of a solution is decreased from pH 8 to pH 6 it means that the concentration of hydrogen ions has increased in the solution.

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3 years ago

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