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frosja888 [35]
3 years ago
5

WHOEVER ANSWERS GETS BRAINLIEST!! (im not lying its facts)

Chemistry
2 answers:
storchak [24]3 years ago
4 0

Answer:

The answer to your question is D. 25 grams.

Assoli18 [71]3 years ago
3 0

Answer

22 grams hope this helps

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What is the kinetic energy of a 25 kg object moving at a velocity of 2.5 m/s?
vovangra [49]

Answer:

<h2>78.13 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

m is the mass

v is the velocity

From the question we have

k =  \frac{1}{2}  \times 25 \times  {2.5}^{2}  \\  = 12.5 \times 6.25 \\  = 78.125 \:  \:  \:  \:  \:  \:  \:  \:

We have the final answer as

<h3>78.13 J</h3>

Hope this helps you

6 0
3 years ago
HELP PLEASE I GIVE 21 POINTS!!!! +BRAINLIEST!!!!!!!
cricket20 [7]
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3 years ago
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The fossil record provides evidence of past life as well as clues to the past climate, environment, and abiotic factors in an ar
murzikaleks [220]
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4 years ago
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Consider the reaction Mg₂Si(s) + 4 H₂O(ℓ) → 2 Mg(OH)₂(aq) + SiH₄(g). How many grams of silane gas (SiH₄) are formed if 25.0 g of
Neko [114]

Answer:

10.60 grams of silane gas are formed.

Explanation:

From the reaction:

Mg₂Si(s) + 4H₂O(l) → 2Mg(OH)₂(aq) + SiH₄(g)          

We know that the limiting reactant is Mg₂Si, so to find the mass of SiH₄ formed we need to calculate the number of moles of Mg₂Si:

\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}}

Where:

m: is the mass of Mg₂Si = 25.0 g

M: is the molar mass of Mg₂Si = 76.69 g/mol

\eta_{Mg_{2}Si} = \frac{m_{Mg_{2}Si}}{M_{Mg_{2}Si}} = \frac{25.0 g}{76.69 g/mol} = 0.33 moles

Now, the stoichiometric relation between Mg₂Si and SiH₄ is 1:1 so:

\eta_{Mg_{2}Si} = \eta_{SiH_{4}} = 0.33 moles

Finally, the mass of SiH₄ is:

m_{SiH_{4}} = \eta_{SiH_{4}}*M_{SiH_{4}} = 0.33 moles*32.12 g/mol = 10.60 g

Therefore, 10.60 grams of silane gas are formed.

I hope it helps you!    

3 0
3 years ago
Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially a
katrin2010 [14]

Answer:

The final temperature of the system is 42.46°C.

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\times (T_f-T_1)=-(m_2\times c\times (T_f-T_2))

where,

c = specific heat of water= 4.18J/g^oC

m_1 = mass of water sample with 100 °C= 50.0 g

m_2 = mass of water sample with 13.7 °C= 100.0 g

T_f = final temperature of system

T_1 = initial temperature of 50 g of water sample= 100^oC

T_2 = initial temperature of 100 g of water =13.7^oC

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

50.0 g\times 4.184 J/g^oC\times (T_f-100^oC)=-(100 g\times 4.184 J/g^oC\times (T_f-13.7^oC))

T_f=42.46^oC

The final temperature of the system is 42.46°C.

5 0
4 years ago
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