BRAINLIEST WILL BE GIVEN!! FAST PLEASE!

There is 200 grams of water on a stove. The water is currently 25 C. How much energy in joules must be used to raise the water t

ad-work [718]

Answer:

Maybe answer is 25

Explanation:

25

6 0

3 years ago

Is an example of the carbon cycle occurring slowly?(1 point)

o-na [289]

Decomposition is a part of the carbon cycle that occurs slowly hence movement of carbon dioxide into the atmosphere when bacteria decomposes dead matter is a slow part of the carbon cycle.

<h3>What is the carbon cycle?</h3>

The carbon cycle is part of the biogeochemical cycles that exist in nature. It refers to the movement of carbon in the ecosystem. The carbon cycle cuts across the air, the land and the water bodies.

The process in the carbon cycle that occurs slowly among the options is the movement of carbon dioxide into the atmosphere when bacteria decomposes dead matter.

Learn more about carbon cycle: brainly.com/question/1627609

7 0

3 years ago

What does acetic anhydride form when it reacts with water?

Stels [109]

What does acetic anhydride form when it reacts with water?

When acetic anhydride reacts with water it forms acetic acid. In other words, The hydrolysis of acetic anhydride in presence of water forms acetic acid.

*Note that:- It is an exothermic reaction.

7 0

3 years ago

Read 2 more answers

Relative formula mass of glucose? (C6H12O6)

Rom4ik [11]

To find the mass of glucose, you must multiply the atomic weight of each of the elements in the molecule by the subscripts in the formula:

$C_{6}=6*(12.01g)=72.06g$

$H_{12}=12*(1.008g)=12.096g$

$O_{6}=6*(15.999g)=95.994g$

Then you add all of them together:

$72.06g+12.096g+95.994g=180.15g$

Therefore, the molar weight of glucose is 180.15 grams.

3 0

3 years ago

The rate of disappearance of HBr in the gas phase reaction 2HBr(g)→H2(g)+Br2(g) is 0.190 Ms−1 at 150 ∘C. The rate of reaction is

nydimaria [60]

Answer:

Rate of reaction is $0.0950M.s^{-1}$

Explanation:

Applying law of mass action for this reaction: $rate=-\frac{1}{2}\frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [H_{2}]}{\Delta t}=\frac{\Delta [Br_{2}]}{\Delta t}$
$-\frac{\Delta [HBr]}{\Delta t}$ represents rate of disappearance of HBr, $\frac{\Delta [H_{2}]}{\Delta t}$ represents rate of appearance of $H_{2}$ and $\frac{\Delta [Br_{2}]}{\Delta t}$ represents rate of appearance of $Br_{2}$
Here, $-\frac{\Delta [HBr]}{\Delta t}=0.190M.s^{-1}$
So, rate of reaction = $\frac{1}{2}\times (0.190M.s^{-1})=0.0950M.s^{-1}$

6 0

3 years ago