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3 years ago

12

Evaporation is ________. check all that apply. check all that apply. an endothermic process sometimes a warming process always a

cooling process sometimes a cooling process an exothermic process always a warming process

1 answer:

Neporo4naja [7]3 years ago

3 0

A cooling process and an exothermic process

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Are plant cells affected when water is scarce

boyakko [2]

<span>Water is vital to the existence of plant life. Not only do plants use it to keep their cells from drying out, but they need water to move nutrients and raw materials throughout their systems to areas where photosynthesis and seed production take place. When water stress occurs, whether caused by drought or root impairment, many invisible processes are affected.</span>

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3 years ago

PLEASE HELPPPP???!!!!!!

sertanlavr [38]

Mass of molecule (g) = Mr of substance over avarogado constant

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3 years ago

Consider this chemical reaction. What chemical is reduced?<br> CH3OH + NAD --> CH2O + NADH

nikklg [1K]

Answer:

CH3OH and NADH

Explanation:

The given chemical reaction is an redox reaction in which reduction and oxidation take place.

In the process of oxidation: electrons are loss while in the process of reduction: electrons are gained.

In the given redox reaction: CH3OH + NAD --> CH2O + NADH

NAD is reduced to NADH as NADH gains one hydrogen electron while CH3OH (methanol) is oxidized to CH2O (methanal) by losing electrons.

So, CH3OH (methanol) and NADH are the reduced forms while NAD and CH2O (methanal) are oxidized forms.

5 0

3 years ago

If the mass of a particular atom is exactly 5 times the mass of an atom of carbon-12, what is its mass

otez555 [7]

The mass is 60 grams.

8 0

3 years ago

A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain

tatiyna

Answer:

$C_m=0.474\frac{J}{g\°C}$

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

$Q_m=-Q_w$

Thus, in terms of masses, specific heats and temperatures we can write:

$m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)$

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

$C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}$

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

$C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}$

Best regards!

6 0

3 years ago