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2 years ago

6

A sample of Krypton gas occupies 66.7 L at 25.0 °C. Assuming constant pressure, what would the temperature of the gas be in kelv

in if the volume increases to 100.0 L?

1 answer:

o-na [289]2 years ago

6 0

Answer:

447

Explanation:

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When 70. milliliter of 3.0-molar Na2CO3 is added to 30. milliliters of 1.0-molar NaHCO3 the resulting concentration of Na+ is 2

tiny-mole [99]

Answer : The resulting concentration of $Na^+$ ion is, 4.5 M

Explanation : Given,

Concentration of $Na_2CO_3$ = $M_1$ = 3.0 M = 3.0 mol/L

Volume of $Na_2CO_3$ = $V_1$ = 70 mL = 0.07 L

Concentration of $NaHCO_3$ = $M_2$ = 1.0 M = 1.0 mol/L

Volume of $NaHCO_3$ = $V_2$ = 30 mL = 0.03 L

First we have to calculate the moles of $Na_2CO_3$ and $NaHCO_3$

$\text{Moles of }Na_2CO_3=\text{Concentration of }Na_2CO_3\times \text{Volume of }Na_2CO_3=3.0mol/L\times 0.07L=0.21mol$

and,

$\text{Moles of }NaHCO_3=\text{Concentration of }NaHCO_3\times \text{Volume of }NaHCO_3=1.0mol/L\times 0.03L=0.03mol$

Now we have to calculate the moles of $Na^+$ ions.

As, 1 mole of $Na_2CO_3$ will give 2 moles of $Na^+$ ions

So, 0.21 moles of $Na_2CO_3$ will give $2\times 0.21=0.42$ moles of $Na^+$ ions

and,

As, 1 mole of $NaHCO_3$ will give 1 mole of $Na^+$ ions

So, 0.03 moles of $NaHCO_3$ will give 0.03 moles of $Na^+$ ions

So,

Total number of moles of $Na^+$ ions = 0.42 + 0.03 =0.45 mole

Total volume of both solution = 70 mL + 30 mL = 100 mL = 0.1 L

Now we have to calculate the concentration of $Na^+$ ions.

$\text{Concentration of }Na^+=\frac{\text{Moles of }Na^+}{\text{Volume of solution}}=\frac{0.45mol}{0.1L}=4.5mol/L=4.5M$

Therefore, the resulting concentration of $Na^+$ ion is, 4.5 M

6 0

2 years ago

Recall that your hypothesis is that these values are the fraction of atoms that are still radioactive after n half-life cycles.

jolli1 [7]

Answer : A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625

Explanation :

Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.

Here n represents the number of half lives.

The amount of substance that remains after n half lives can be calculated using the given formula, $0.5^{n}$

So when we have n =1,

Fraction of substance that remains = 0.5¹ = 0.5.

That means after first half life over, the amount of substance that remains is 0.5 times that of original.

Therefore we have A = 0.5

When n = 2, we have 0.5² = 0.25

So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original

Therefore B = 0.25

When n = 3, we have 0.5³ = 0.125

So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.

Therefore we have C = 0.125

When n = 6 , we have 0.5⁶ = 0.015625

So D = 0.015625

When n = 8, we have 0.5⁸ = 0.00390625

Therefore E = 0.00390625

The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.

8 0

2 years ago

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Old photographic flashbulbs burn magnesium metal in a reaction that causes a flash of brilliant white light. A photographer woul

OLga [1]

Answer is: A. Chemical energy to electromagnetic energy and thermal energy.

Balanced chemical reaction: 2Mg(s) + O₂(g) → 2MgO(s) + energy.

This is chemical change (chemical reaction), because new substance (magnesium oxide MgO) is formed, the atoms are rearranged and the reaction is followed by an energy change (exothermic reaction because energy is released).

Chemical changes (chemical synthesis) is when a substance combines with another (in this example magnesium and oxygen) to form a new substance.

5 0

2 years ago

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A carpet sells for 27.99 a square yard. What is the price of the carpet square per square meter ?

klio [65]

There are 9 square meters in one square yard so divide 27.99 by 9

27.99/9=3.11

one square meter is $3.11

I hope I've helped!

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2 years ago

Suppose that magnesium would react exactly the same as copper in this experiment. how many grams of magnesium would have been us

Vikentia [17]

The solution for this problem would be:

We are looking for the grams of magnesium that would have been used in the reaction if one gram of silver were created. The computation would be:

1 g Ag (1 mol Mg) (24.31 g/mol) / (2mol Ag)(107.87g/mol) = 0.1127 grams of Magnesium

6 0

2 years ago