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balandron [24]
3 years ago
13

8.310x10^2 – 7.210x10^1[?]x10^[?]​

Chemistry
1 answer:
alisha [4.7K]3 years ago
7 0

Answer:

7.589\times 10^{2}

Explanation:

The expression can be solved mathematically as follows:

1) 8.310\times 10^{2}-7.210\times 10^{1} Given

2) 8.310\times 10^{1+1} - 7.210\times 10^{1} Definition of sum

3) (8.310\times 10^{1})\times 10^{1}-7.210\times 10^{1} a^{m+n} = a^{m}\cdot a^{n}/Associative property

4) (8.310\times 10^{1}-7.210)\times 10^{1} Distributive property

5) (83.100-7.210)\times 10^{1} Multiplication

6) 75.89\times 10^{1} Subtraction.

7) (7.589\times 10^{1})\times 10^{1} Multiplication/Associative property

8) 7.589\times (10^{1}\times 10^{1}) Associative property

9) 7.589\times 10^{2} a^{m+n} = a^{m}\cdot a^{n}/Result

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If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
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<u>Answer:</u> The final temperature of the mixture is  49°C

<u>Explanation:</u>

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\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

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Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

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The equation used to calculate heat released or absorbed follows:

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m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 120 g

m_2 = mass of cold water = 230 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of hot water = 95°C

T_2 = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]

T_{final}=49^oC

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