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klemol [59]
3 years ago
7

HELP PLEASE!!!!!!!!! What if you started with 100 grams of wood and only have 95 grams of ashes produced, is mass still conserve

d? If so where is the missing 5 grams?
Chemistry
1 answer:
Lerok [7]3 years ago
7 0
My guess is smoke. Smoke is matter.
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What is the state of water at room temperature? Please someone help thank you
velikii [3]

it is in a liquid state

in room temperature

7 0
3 years ago
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A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
How many particles are present in 8.0 moles of silver?
Zielflug [23.3K]

Answer:the answer is b

Explanation:I took the test and got it right

3 0
3 years ago
A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release
Marrrta [24]

Answer:

V_2=12.1L

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:

T_1=20+273=293K\\\\T_2=-40+273=233K

Then, we obtain:

V_2=\frac{P_1V_1T_2}{T_1P_2}\\\\V_2=\frac{0.987atm*8.50L*233K}{293K*0.550atm}\\\\V_2=12.1L

Best regards!

5 0
3 years ago
A sealed can of your favorite soda has a carbon dioxide gas (C0) volume of 0.05 L. When it is refrigerated, the
liubo4ka [24]

Answer:

<h2>0.102 L</h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{207000 \times 0.05}{101000}  =  \frac{10350}{101000}  \\  = 0.102475...

We have the final answer as

<h3>0.102 L</h3>

Hope this helps you

8 0
3 years ago
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