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3 years ago

Why does opening the air valve of a tire at a constant temperature decrease the pressure

1 answer:

n200080 [17]3 years ago

5 0

Air pressure decreases along with the temperature. It is due to the molecular motion of gas particles. The gas particles decrease as the temperature decreases.

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Sodium metal reacts with water to produce hydrogen gas according to the following equation: 2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) Th

Maru [420]

Answer:

Number of moles of sodium reacted = 0.707 moles

Explanation:

P(H₂) = P(T) – P(H₂O)

P(H₂) = 754 – 17.5 = 736.5 mm Hg

Use the ideal gas equation which

PV= nRT, where P is the pressure V is the volume, n is the number of moles R is the Gas Constant and T is temperature

Re- arrange to calculate the number of moles and using the data provided

n = P x V/R x T

n =736.5 x 8.77/62.36367 x (mmHg/mol K) x (20 + 273)

n = 0.35348668

n = 0.353 moles H₂

from the equation we know that

0.353 mole H₂ x 2mole Na/1mole H₂, So

0.353 x 2 = 0.707 mole Na

The number of moles of Sodium metal reacted were 0.707 moles.

3 0

3 years ago

Calculate the mass percent of calcium chloride in 8.87 g of calcium chloride in 65.1 g of water

zhannawk [14.2K]

Answer:

11.99 % ≅ 12.0%.

Explanation:

∵ mass % = [mass of solute/mass of solution] x 100.

mass of solute (CaCl₂) = 8.87 g & mass of solution = 8.87 g + 65.1 g = 73.97 g.

∴ mass % of (CaCl₂) = [mass of solute/mass of solution] x 100 = ( 8.87 g/ 73.97 g) x 100 = 11.99 % ≅ 12.0%.

7 0

3 years ago

. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:

mafiozo [28]

Answer:

See explanation

Explanation:

a) The equation of the reaction is;

2Na + Cl2 ------>2NaCl

Number of moles of sodium = 10g/23 g/mol = 0.43 moles

If 2 moles of sodium reacts with 1 mole of Cl2

0.43 moles reacts with 0.43 * 1/2 = 0.215 moles of Cl2

Mass = 0.215 moles of Cl2 *71 g/mol = 15.265 g

b) Equation of the reaction;

HgO -> Hg + O2

1.252 moles of HgO 1.252/32 gmol = 0.039 moles

1 mole of HgO yields 1 mole of oxygen hence

0.039 moles of HgO yields 0.039 moles of oxygen

Mass of oxygen = 0.039 moles * 32 g/mol = 1.248 g

c) Equation of the reaction;

2NaNO3 -----> 2NaNO2 + O2

Number of moles of 128 g of oxygen = 128g/32 g/mol = 4 moles

2 moles of NaNO3 yields 1 mole of oxygen

x moles of NaNO3 yields 4 moles of oxygen

x = 8 moles of NaNo3

Mass of NaNO3 = 8 * 85 g/mol = 680 g of NaNo3

6 0

3 years ago

In the following reaction, how many grams of NaBr will produce 244 grams of NaNO3? Pb(NO3)2(aq) 2 NaBr(aq) PbBr2(s) 2 NaNO3(aq)

lyudmila [28]

Answer : The mass of $NaBr$ is, 295.323 grams

Solution :

First we have to calculate the moles of $NaNO_3$ .

$\text{Moles of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Molar mass of }NaNO_3}=\frac{244g}{85g/mole}=2.87moles$

Now we have to calculate the moles of NaBr.

The balanced chemical reaction is,

$Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)$

From the balanced reaction we conclude that

As, 2 moles of $NaNO_3$ react with 2 moles of $NaBr$

So, 2.87 moles of $NaNO_3$ react with 2.87 moles of $NaBr$

Now we have to calculate the mass of NaBr.

$\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr$

$\text{Mass of }NaBr=(2.87mole)\times (102.9g/mole)=295.323g$

Therefore, the mass mass of $NaBr$ is, 295.323 grams

7 0

3 years ago

What is the concentration of OH − and pOH in a 0.00066 M solution of Ba ( OH ) 2 at 25 ∘ C? Assume complete dissociation.

Allushta [10]

Answer: The hydroxide ion concentration and pOH of the solution is $1.32\times 10^{-3}M$ and 2.88 respectively

Explanation:

We are given:

Concentration of barium hydroxide = 0.00066 M

The chemical equation for the dissociation of barium hydroxide follows:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M$

Putting values in above equation, we get:

$pOH=-\log(1.32\times 10^{-3})\\\\pOH=2.88$

Hence, the hydroxide ion concentration and pOH of the solution is $1.32\times 10^{-3}M$ and 2.88 respectively

3 0

3 years ago