An unknown substance with a mass of 83 grams absorbs 1,108 J while undergoing a
temperature increase of 11 °C. What is the specific heat of the substance?

Look at picture
ASAP please will mark as brainlist don’t got much time

An Unknown Substance With A Mass Of 83 Grams Absorbs 1,108 J While Undergoing Atemperature Increase Of

Answers

Answer 1

Answer:

The specific heat of the substance is 1.21 [tex]\frac{J}{g*C}[/tex]

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:

Q = c*m*ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

Q=1108 J c= ? m= 83 g ΔT= 11 C

Replacing:

1108 J= c*83 g* 11 C

Solving:

[tex]c=\frac{1108 J}{83 g* 11 C}[/tex]

c= 1.21 [tex]\frac{J}{g*C}[/tex]

The specific heat of the substance is 1.21 [tex]\frac{J}{g*C}[/tex]


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it would be nitrogen
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i hope that's correct

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To prepare 800mL of 1M phosphate buffer of pH 7.2 (PKa=7.1), by using salt = K2HPO4 and acid = KH2PO4, the quantity of salt required is

Answers

Answer:

78.04g of 0.448 moles must be added

Explanation:

Using H-H equation we can find the pH of the buffer:

pH = pKa + log [A⁻] / [HA]

Where pH is the pH of the buffer = 7.2

pKa = 7.1

[A⁻] = [K₂HPO₄]

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0.1 = log [K₂HPO₄] / [KH₂PO₄]

1.2589 = [K₂HPO₄] / [KH₂PO₄] (1)

And as the  concentration of the buffer is:

1M = [K₂HPO₄] + [KH₂PO₄] (2)

Replacing (2) in (1):

1.2589 = 1M - [KH₂PO₄] / [KH₂PO₄]

1.2589 [KH₂PO₄] = 1M - [KH₂PO₄]

2.2589 [KH₂PO₄] = 1M

[KH₂PO₄] = 0.44M

And [K₂HPO₄] = 0.56M

In 800mL = 0.8L:

0.8L * (0.56mol / L) = 0.448 moles K₂HPO₄. The mass is -Molar mass K₂HPO₄: 174.2g/mol-:

0.448 moles * (174.2g / mol) =

78.04g of 0.448 moles must be added

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Look at picture
ASAP please will mark as brainlist don’t got much time

Answers

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