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3 years ago

How many sodium ions are contained in 99.6 mg of na2so3? The molar mass of na2so3 is 126.05 g/mol.

Chemistry

2 answers:

marin [14]3 years ago

8 0

The answer is D shadowing

Norma-Jean [14]3 years ago

3 0

Answer:

9.52x10²⁰ sodium ions

Explanation:

First, we need to convert the mass in moles (n)

n = mass/molar mass

The moalr mass is given in g/mol, so let's transform the mass in gram:

1 g ------------ 1000 mg

x g------------- 99.6 mg

By a direct simple three rule:

1000x = 99.6

x = 0.0996 g

n = 0.0996/126.05

n = 7.90x10⁻⁴ mol of Na₂SO₃

The ionization reaction is:

Na₂SO₃ → 2Na⁺ + SO₃⁻

So, for 1 mol of Na₂SO₃, will be two moles of Na⁺ (sodium ion), then:

1 mol of Na₂SO₃ --------------- 2 moles of Na⁺

7.90x10⁻⁴ ------------------------- x

By a simple direct three rule:

x = 1.58x10⁻³ mol of Na⁺

Avogadro has related the number of moles with the number of the matter: atoms, ions, molecules. So, 1 mol has 6.02x10²³ ions.

1 mol ------------------- 6.02x10²³

1.58x10⁻³ mol ----------- x

By a direct simple three rule:

x = 9.52x10²⁰ sodium ions

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A 25. 0-ml sample of 0. 150 m hydrocyanic acid is titrated with a 0. 150 m naoh solution. What is the ph beforeany base is added

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Answer:

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Explanation:

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3 years ago

A 250 mL sample of gas is collected over water at 35°C and at a total pressure of 735 mm Hg. If the vapor pressure of water at 3

ELEN [110]

Answer:

The volume of the gas sample at standard pressure is 819.5ml

Explanation:

Solution Given:

let volume be V and temperature be T and pressure be P.

$V_1=250ml$

$V_2=?$

$P_{total}=735 mmhg$

1 torr= 1 mmhg

42.2 torr=42.2 mmhg

so,

$P_{water}=42.2mmhg$

$T_1=35°C=35+273=308 K$

Now

firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.

$P_{gas}=P_{total}-P_{water}$

=735-42.2=692.8 mmhg

Now

By using combined gas law equation:

$\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}$

$V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1$

$V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1$

Here $P_2 \:and\: T_2$ are standard pressure and temperature respectively.

we have

$P_2=750mmhg \:and\: T_2=273K$

Substituting value, we get

$V_2=\frac{692.8}{750}*\frac{273}{308} *250$

$V_2= 819.51 ml$

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2 years ago

What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Vadim26 [7]

Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated from 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
c= 0.4494 J/g°C
ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$