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3 years ago

What is the molarity of a solution that contains 4.53 moles of lithium nitrate in 2.85 liters of solution

Chemistry

2 answers:

scZoUnD [109]3 years ago

7 0

Answer:

can i have brainlest

Explanation:

Colt1911 [192]3 years ago

3 0

Molarity = moles÷liters

4.53 moles ÷ 2.85 liters = 1.59 M

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The scale most widely used by scientist for measuring earth quakes is called what

DanielleElmas [232]

A sesimograph or a sesimogram.

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3 years ago

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What is the percent by volume of ethanol (C2H60, or ethyl alcohol) in the final solution when 90 mL of ethanol is diluted to a v

r-ruslan [8.4K]

Answer:

16.0%.

Explanation:

Volume percent of a substance is the ratio of the substance volume to the solution volume multiplied by 100.

V % of ethanol = (volume of ethanol / volume of the solution) x 100.

volume of ethanol = 90.0 mL, volume of the solution = 550.0 mL.

∴ V % of ethanol = (90.0 mL / 550.0 mL) x 100 = 16.36% ≅ 16.0%.

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3 years ago

At STP how many molecules of nitrogen gas are in 224 L?

NeTakaya

Answer:

6.022×10²² molecules

Explanation:

Given data:

Volume of nitrogen = 224 L

Pressure = standard = 1 atm

Temperature = standard = 273 K

Number of molecules = ?

Solution:

PV = nRT

1 atm × 224 L = n × 0.0821 atm.L/mol.K × 273 K

224 atm.L = n ×22.41 atm.L/mol

n = 224 atm.L/22.41 atm.L/mol

n = 10 mol

1 mole contain 6.022×10²³ molecules

10 mol×6.022×10²³ molecules/ 1 mol

60.22×10²³ molecules

6.022×10²² molecules

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3 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]

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

<u>Explanation:</u>

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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4 years ago

How many electrons are represented by a dash (-) in a lewis dot structure compound?

vivado [14]

A. 2
The dash represents two electrons

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3 years ago