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LenKa [72]
3 years ago
13

How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat

Chemistry
1 answer:
inna [77]3 years ago
8 0

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C    

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

  =670*4.1*40.3

=110704.1

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Which is best represented by the equation?
sergeinik [125]
A- law of conservation of energy
3 0
3 years ago
SOMEONE PLS HELP
Gekata [30.6K]

Answer:

<h2>464.85 mL</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we're finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

100.7 kPa = 100,700 Pa

95.1 kPa = 95,100 Pa

We have

V_2 =  \frac{100700 \times 439}{95100}  =  \frac{44207300}{95100}  \\  = 464.8506...

We have the final answer as

<h3>464.85 mL</h3>

Hope this helps you

6 0
3 years ago
Where is most of Earth's liquid water ? what must be done so humans can drink it ?
LiRa [457]
You're right it's the oceans.. since most is salt water, the salt has to be boiled out 
4 0
3 years ago
Read 2 more answers
Please help I will give brainliest!
Fiesta28 [93]

Answer:

b, H2O(s) r H2O(g)

Explanation:

entropy is heat, so increase in heat would mean water gets evaporated or melted, or both in this case. so the only choice above that showed increase in heat is from solid(ice) to gas(water vaper) due to increase in heat in the reaction.

7 0
3 years ago
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO
Irina-Kira [14]

Answer:

Amount of excess Carbon (ii) oxide left over = 23.75 g

Explanation:

Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

Molar mass of Fe₂O₃ = 160 g/mol;

Molar mass of Carbon (ii) oxide = 28 g/mol

From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g)  of carbon (ii) oxide

450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide

Therefore the excess reactant is carbon (ii) oxide.

Amount of excess Carbon (ii) oxide left over = 260 - 236.25

Amount of excess Carbon (ii) oxide left over = 23.75 g

5 0
3 years ago
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