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2 years ago

A 12.0g sample of a substance requires 50.9 J to raise its temperature by 6.00 c. What is it’s specific heat

Chemistry

1 answer:

horsena [70]2 years ago

7 0

$\qquad\qquad\huge\underline{{\sf Answer}}$

Let's find the specific heat of sample using given formula :

$\qquad \tt \dashrightarrow \:Q = m \: s \triangle T$

where,

Q = heat transfer

m = mass

S = specific heat

$\tt \triangle T$ = change in temperature

$\qquad \tt \dashrightarrow \:50.9 = 12 \times \: s \times 6$

$\qquad \tt \dashrightarrow \:s = \dfrac{50.9}{72}$

$\qquad \tt \dashrightarrow \:s = 0.707 \:$

Therefore, specific heat of that substance is 0.707 J/g°C

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7. How many significant figures are in the following numbers?

Vlad1618 [11]

There are
A. 6 significant figures
B.5 sig figs
C.2 sig figs
D.5 sig figs

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3 years ago

What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0

expeople1 [14]

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = 45.1mL

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = $\frac{[OH^-][HNO_2]}{[NO_2]}$

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

Where X is reaction coordinate

Replacing in Kb:

1.41x10⁻¹¹ = $\frac{[X][X]}{[0.211 -X]}$

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ FALSE ANSWER. There is no negative concentrations.

X = 1.72x10⁻⁶. Right answer.

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; pH = 8.24

3 0

3 years ago

The densities of gases a, b, and c at stp are 1.25 g/l, 2.86 g/l and 0.714 g/l, respectively. calculate the molar mass of each s

kicyunya [14]

The molar mass of a, b and c at STP is calculated as below

At STP T is always= 273 Kelvin and ,P= 1.0 atm

by use of ideal gas equation that is PV =nRT
n(number of moles) = mass/molar mass therefore replace n in the ideal gas equation

that is Pv = (mass/molar mass)RT
multiply both side by molar mass and then divide by Pv to make molar mass the subject of the formula

that is molar mass = (mass x RT)/ PV

density is always = mass/volume

therefore by replacing mass/volume in the equation by density the equation
molar mass=( density xRT)/P where R = 0.082 L.atm/mol.K

the molar mass for a
= (1.25 g/l x0.082 L.atm/mol.k x273k)/1.0atm = 28g/mol

the molar mass of b
=(2.86g/l x0.082L.atm/mol.k x273 k) /1.0 atm = 64 g/mol

the molar mass of c

=0.714g/l x0.082 L.atm/mol.K x273 K) 1.0atm= 16 g/mol

therefore the
gas a is nitrogen N2 since 14 x2= 28 g/mol
gas b =SO2 since 32 +(16x2)= 64g/mol
gas c = methaneCH4 since 12+(1x4) = 16 g/mol

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2 years ago

The net ionic equation for the reaction between aqueous nitric acid and aqueous sodium hydroxide is ________. h+ (aq) + na+ (aq)

Nana76 [90]

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7 0

2 years ago

Oxygen decays to form nitrogen.

tatyana61 [14]

Answer:

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3 years ago