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3 years ago

POINTSSS!!!!!!

1 answer:

4 0

1 mole CO₂----------- 6.02x10²³ molecules
4 moles CO₂ ---------- ??

4 x ( 6.02x10²³) / 1 =

= 2.41 x 10²⁴ / 1 => 2.41 x 10²⁴ molecules of CO₂

Answer C

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What are the factors affecting qualitative analysis in chemistry

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Answer:

Techniques and Tests

Qualitative analysis typically measures changes in color, melting point, odor, reactivity, radioactivity, boiling point, bubble production, and precipitation.

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1 year ago

What is the volume of 300. g of mercury vapor at 822K and 0.900 atm?

Dominik [7]

Answer:

112.2L

Explanation:

Volume (V) = 300g

Temperature (T) = 822K

Pressure (P) = 0.9atm

using the ideal gas equation;

$PV = nRT\\\\ V = \frac{nRT}{P}$

Molar gas constant (R) = $0.0821L.atm/mol.K$

Mole (n) = $\frac{Mass (m)}{Molar mass (M)}$ Molar mass of Mercury = 200.59g/mol

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2 years ago

"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the

soldier1979 [14.2K]

Answer: The specific heat of calorimeter is 30.68 J/g°C

Explanation:

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of hot water = 46.7 g

$m_2$ = mass of cold water = 45.33 g

$T_{final}$ = final temperature = 59.4°C

$T_1$ = initial temperature of hot water = 80.6°C

$T_2$ = initial temperature of cold water = 40.6°C

$c_1$ = specific heat of hot water = 4.184 J/g°C

$c_2$ = specific heat of cold water = 4.184 J/g°C

$c_3$ = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

$46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)$

$c_3=30.68J/g^oC$

Hence, the specific heat of calorimeter is 30.68 J/g°C

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3 years ago

if 0.3 moles of carbon reacts completely with 0.3 moles of molecular oxygen, what is the yield in Percent of 6.6 g of CO2 are fo

Anettt [7]

Answer is:the yield is 50%.

Chemical reaction: C + O₂ → CO₂.
n(C) = 0.3 mol; amount of substance.
n(O₂) = 0.3 mol.
From chemical reaction: n(C) : n(CO₂) = 1 : 1.
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m(CO₂) = n(CO₂) · M(CO₂).
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3 years ago

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Answer:

Explanation:

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2 years ago

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