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bija089 [108]
3 years ago
13

The specific heat of gold is 0.031 calories/gram°C and the specific heat of silver is 0.057 calories/gram°C. If equal amounts of

each metal are exposed to equal heating, which will heat up faster?
Chemistry
2 answers:
OLga [1]3 years ago
8 0

Answer : Gold will heat up faster.

Explanation : Given,

The Specific heat of Silver and gold are 0.057 & 0.0314 calories/gram^{0}C respectively.

Specific heat is the amount of energy/heat required to raise the temperature of 1 Kg mass by 1 Kelvin.

The formula of Specific heat is,

q=m\times c\times (T_{final}-T_{initial})

Where,

c = Specific heat

q = Heat required

T_{final} = final temperature

T_{initial} = initial temperature

we are taking initial temperature as zero. According to the question, the mass and the heat are same.

Now put all the values in above formula, we get

For gold,

q=m\times 0.031\times (T_{gold}-0)        .........(1)

For silver,

q=m\times 0.057\times (T_{silver}-0)      ......... (2)

Now equating equation (1) and (2), we get

T_{gold}=1.83\times T_{silver}

From this, we conclude that the temperature of gold is 1.83 times higher than the temperature of silver.

Therefore, Gold will heat up faster.  

Vikki [24]3 years ago
3 0
I think the substance that will heat up faster would be the silver metal since it has a higher heat capacity. Heat capacity is the amount of heat needed to raise the temperature of the system into one degree. Heat capacity and heat energy is directly related so higher value of heat capacity would lead to higher heat energy.
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zmey [24]

Your answer will be B because sodium and chlorine and both elements and two elements combine to make a compound

7 0
3 years ago
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How many Liters of space will a 70.0g sample of CO2 occupy?
Tanzania [10]

Answer:

  35.6 liters at STP

Explanation:

The molar mass of carbon dioxide is about 44.01 g/mol. The volume of a mole of ideal gas at STP is 22.4 L, so the volume of 70.0 g will be ...

  (70.0g)/(44.01 g/mol)·(22.4 L/mol) ≈ 35.6 L

5 0
3 years ago
What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.
AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

pH=pK_{a} + log(\frac{[Base]}{[Acid]})

According to the given conditions, the equation will become as follow

pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

pH=6.86 + log(\frac{0.058}{0.042})

pH=7.00

5 0
4 years ago
Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.
BaLLatris [955]

Answer:

59.8%

Explanation:

First find the Mr of manganese (III) nitrate.

Mr of Mn(NO₃)₃ = 54.9 + (14 × 3) + (16 × 3 × 3) = <u>240.9</u>

Since we have to find the percentage composition of oxygen, we need to find the Mr of oxygen in the compound, which is:

Mr of (O₃)₃ = (16 × 3) × 3 = <u>144</u>

Now we can find percentage composition / percentage by mass of oxygen.

% composition = \frac{Mr\ of\ oxygen\ in\ compound}{Mr\ of\ compound} × 100

% composition = \frac{144}{240.9} × 100 = <u>59.776%</u>

∴ % compostion of oxygen in maganese(III)nitrate is 59.8% (to 3 significant figures).

8 0
3 years ago
The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?
Vitek1552 [10]
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
     pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
     [OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
     NaA → Na+ + A- 
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
                             A- + H2O ⇌ HA + OH-
     Initial             0.0100            0       0
     Change        -x                    +x     +x
     Equilibrium    0.0100-x         x       x

We can now calculate the Kb for A-:
     Kb = [HA][OH-] / [A-] 
           = x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
     Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
     Ka = Kw / Kb 
           = 1.00x10^-14 / 1.00x10^-4
           = 1.00x10^-10
7 0
3 years ago
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