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aleksandrvk [35]
3 years ago
5

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Chemistry
2 answers:
balandron [24]3 years ago
8 0

Answer:

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Explanation:

Andreas93 [3]3 years ago
6 0

Answer:

sure okay I'll join your game

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What volume of lead (of density 11.3 g/cm3 ) has the same mass as 395 cm3 of a piece of redwood (of density 0.38 g/cm3 )? Answer
Zepler [3.9K]
d_{1}=\frac{m}{V_{1}}\\\\
V_{1}=395cm^{3}\\
d_{1}=0,38\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ m=395cm^{3}*0,38\frac{g}{cm^{3}}=150,1g\\\\\\
d_{2}=\frac{m}{V_{2}}\\\\
m=150,1g\\
d_{2}=11,3\frac{g}{cm^{3}} \ \ \ \ \Rightarrow \ \ \ \ V_{2}=\frac{150,1g}{11,3\frac{g}{cm^{3}}}\approx13,28cm^{3}
8 0
4 years ago
How do scientist communicate the results of an investigation?​
maks197457 [2]

Answer:

scientists often communicate their research results in three general ways:

1) One is to publish their results in peer-reviewed journals that can be ready by other scientists.

2) Two is to present their results at national and international conferences where other scientists can listen to presentations

Explanation:

5 0
3 years ago
Read 2 more answers
A 52.0 g of Copper (specific heat=0.0923cal/gC) at 25.0C is warmed by the addition of 299 calories of energy. find the final tem
Leto [7]

Answer : The final temperature of the copper is, 87.29^oC

Solution :

Formula used :

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

where,

Q = heat gained  = 299 cal

m = mass of copper = 52 g

c = specific heat of copper = 0.0923cal/g^oC      

\Delta T=\text{Change in temperature} 

T_{final} = final temperature = ?

T_{initial} = initial temperature = 25^oC

Now put all the given values in the above formula, we get the final temperature of copper.

299cal=52g\times 0.0923cal/g^oC\times (T_{final}-25^oC)

T_{final}=87.29^oC

Therefore, the final temperature of the copper is, 87.29^oC

4 0
4 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
Vinegar, fruit juice and cola are examples of
andrew-mc [135]

Answer:

strong bases

Explanation:

7 0
3 years ago
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