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3 years ago

Would a material with a higher specific heat cool faster than a material with a lower specific heat?

Chemistry

2 answers:

MaRussiya [10]3 years ago

6 0

Yes because the specific heat your multiplying by would be higher

Mandarinka [93]3 years ago

5 0

Yes because the heat would be higher and doubled from the heat of that that is in the lower heat material

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an element has 8 protons and 8 nuetrons in its nucleus. what is the atomic number and atomic mass of this element

Xelga [282]

Answer:

Atomic number= 8

Atomic mass= 16

Explanation:

You can determine the atomic number by counting the number of protons. Atomic mass is determined by adding the protons and neutrons together.

Hope this helps!

5 0

3 years ago

0 ml of a 1.20 m solution is diluted to a total volume of 228 ml. a 114-ml portion of that solution is diluted by adding 111 ml

ozzi

54.0 ml of a 1.2 m solution was diluted to a total volume of 228 ml
USing dilution equation
M1V1 = M2V2
Where M1V1 is before dilution while M2V2 is after dilution.
Therefore;
M1= 1.20 M, V2= 54 ml,
M2= V2 = 228 ml

1.20 M × 54 ml = M2 × 228 ml
M2 = (1.2 ×54)/ 228 ml
= 0.2842 M

0.2842 × 114 ml = M2 × (114 +111)
M2 = (0.2842 ×114)/ 225
= 0.143995
= 0.144 M

3 0

3 years ago

The rate of disappearance of HBr in the gas phase reaction 2HBr(g)→H2(g)+Br2(g) is 0.190 Ms−1 at 150 ∘C. The rate of reaction is

nydimaria [60]

Answer:

Rate of reaction is $0.0950M.s^{-1}$

Explanation:

Applying law of mass action for this reaction: $rate=-\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [H_{2}]}{\Delta t}=\frac{\Delta [Br_{2}]}{\Delta t}$
$-\frac{\Delta [HBr]}{\Delta t}$ represents rate of disappearance of HBr, $\frac{\Delta [H_{2}]}{\Delta t}$ represents rate of appearance of $H_{2}$ and $\frac{\Delta [Br_{2}]}{\Delta t}$ represents rate of appearance of $Br_{2}$
Here, $-\frac{\Delta [HBr]}{\Delta t}=0.190M.s^{-1}$
So, rate of reaction = $\frac{1}{2}\times (0.190M.s^{-1})=0.0950M.s^{-1}$

6 0

3 years ago

Enter the maximum number of electrons in each type of sublevel (s, p, d, and f, respectively)

Alika [10]

S sub-levels contain a maximum of 2 electrons.
p sub-levels contain a maximum of 6 electrons.
d sub-levels contain a maximum of 10 electrons.
<span>f sub-levels contain a maximum of 14 electrons. </span>

6 0

3 years ago

Calculate the heat required to form a liquid solution at 1356 k starting with 1 mole of cu and 1 mole of ag at 298 k. at 1356 k

tatyana61 [14]

Answer:

$Q=87924J$

Explanation:

Hello,

At first, we must consider that at the beginning, both copper and silver are solid, so we must melt both of them, thus, the following enthalpic routes are proposed in order to know each heat because coppers melting points as a value of 1356K and silver's melting point 1235K:

$H_{Cu}=n_{Cu}*(Cp_{solid,Cu}(T_2-T_1)+H_{melt})\\H_{Cu}=1mol*(0.3768J/gK*63.546g/mol*(1356-298)K+13590J/mol) \\H_{Cu}=38922.89J$

$H_{Ag}=n_{Ag}*(Cp_{solid,Ag}(T_m-T_1)+H_{melt}+Cp_{liquid,Ag}(T_2-T_m))\\H_{Ag}=1mol*(0.235J/gK*107.868g/mol*(1235-298)K+11300J/mol+0.28J/gK*107.868g/mol*(1356-1235)K) \\H_{Ag}=38706.56J$

Now, the released heat due to the mixing process turns out into:

$H_{mix}=2mol*(20590x_{Cu}x_{Ag}J/mol)\\H_{mix}=2mol*20590(0.5)(0.5)J/mol\\H_{mix}=10295J$

Finally, the total heat required is found by:

$Q=H_{Cu}+H_{Ag}+H_{mix}\\Q=38922.89J+38706.56J+10295J\\Q=87924J$

Best regards.

3 0

4 years ago