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Anuta_ua [19.1K]
3 years ago
6

6. What is the total number of calories lost when 10

Chemistry
1 answer:
Nadusha1986 [10]3 years ago
3 0

Answer:

B

Explanation:

so you started out with 80

you have to cool it till 60

so it is decreasing which means you have to subtract

Note=if something is increasing or wanting to go up its addition but for this problem 60 is less then 80 so we are doing subtraction

80-60=20

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Kahdbiaddsk;wvfodas vs;oi vsd;iv sdv;sidv s;vowdbvw;dichvbwdv;iwebfwd;iwdbcv;wkfbsdliasbv;soqivbasd
storchak [24]

Answer:

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6 0
3 years ago
Read 2 more answers
1 L, 1000 mL, and 1000 cm are considered to be<br><br> EQUIVALENT or NON EQUIVALENT UNITS.
antoniya [11.8K]

Answer:

equivalent

Explanation:

hope it works

7 0
2 years ago
What is the coefficient of h2 when you balance the equation for this redox reaction? no2(g) + h2(g) → nh3(g) + h2o(l)?
laiz [17]
To balance any redox equation, first divide the reaction into two half reactions, and balance the atom whose oxidation number is changing: 

<span>NO2 --> NH3 </span>
<span>H2 --> H2O </span>

<span>Next, balance oxygens by adding H2O where needed, and then balance hydrogen by adding H+: </span>

<span>NO2 --> NH3 + 2 H2O </span>
<span>H2O + H2 --> H2O </span>


<span>7 H+ + NO2 --> NH3 + 2 H2O </span>
<span>H2O + H2 --> H2O + 2 H+ </span>

<span>Next, balance charges by adding electrons (e-): </span>

<span>7 H+ + NO2 + 7 e- --> NH3 + 2 H2O </span>
<span>H2O + H2 --> H2O + 2 H+ + 2 e- </span>

<span>Now, multiply one or both equations by small numbers to make the number of electrons the same in the two half reactions. In this case, multiply the top equation by 2 and the bottom one by 7 to give you 14 electrons in each half reaction: </span>

<span>14 H+ + 2 NO2 + 14 e- --> 2 NH3 + 4 H2O </span>
<span>7 H2O + 7 H2 --> 7 H2O + 14 H+ + 14 e- </span>

<span>Finally, add the two half reactions and simplify by cancelling anything that is on both sides. This will give you the final balanced reaction: </span>

<span>2 NO2 + 7 H2 --> 2 NH3 + 4 H2O </span>

<span>So, the answer to the original question is (A) 7
</span>
5 0
4 years ago
What are the effects of inhalation of acetone vapor?
Ipatiy [6.2K]

Answer:

it can cause irritation in eyes throat and lungs and u can become lightheaded, or possibly fall in a coma

5 0
3 years ago
370 cm3 of water at 80°C is mixed with 120 cm3 of water at 4°C. Calculate the final equilibrium temperature, assuming no heat is
NikAS [45]

Answer : The final temperature of the mixture is 61.4^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)

where,

c_1 = c_2 = specific heat of water = same

m_1 = m_2 = mass of water  =  same

\rho_1 = \rho_2 = density of water = 1.0 g/mL

V_1 = volume of water at 80.0^oC  = 370cm^3=370mL

V_2 = volume of water at 4^oC  = 120cm^3=120mL

T_f = final temperature of mixture = ?

T_1 = initial temperature of water = 80.0^oC

T_2 = initial temperature of water = 4^oC

Now put all the given values in the above formula, we get:

(\rho_1\times V_1)\times (T_f-T_1)=-(\rho_2\times V_2)\times (T_f-T_2)

(1.0g/mL\times 370mL)\times (T_f-80.0)^oC=-(1.0g/mL\times 120mL)\times (T_f-4)^oC

T_f=61.4^oC

Therefore, the final temperature of the mixture is 61.4^oC

6 0
4 years ago
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