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3 years ago

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:

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

Hope this helps you

6 0

3 years ago

HELP PLEASE I GIVE 21 POINTS!!!! +BRAINLIEST!!!!!!!

cricket20 [7]

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7 0

3 years ago

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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