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3 years ago

A bracelet containing 0.364 mol of metal atoms is 66% gold. How many particles of gold atoms are in the bracelet?

Chemistry

1 answer:

Katyanochek1 [597]3 years ago

7 0

Answer:

129.9g

Explanation:

196.966569 * 0.66 = 129.9g

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How many calories of heat are required to raise the temperature of 1.00kg of water from 10.2 degrees Celsius to 26.8 degrees Cel

Vika [28.1K]

Answer:

Explanation:

we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly

Q=mcΔT

ΔT=T2-T1

ΔT=26.8-10.2=16.6

C for water is 4.184

therefore

Q=1.00*4.184*16.6

Q=69.4 j

now we have to covert joule into calorie

1 calorie =4.2 j

x calorie=69.4 j/2

so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius

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A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24

Setler79 [48]

Answer:

For Part A: The partial pressure of Helium is 218 mmHg.

For Part B: The mass of helium gas is 0.504 g.

Explanation:

For Part A:

We are given:

$p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg$

To calculate the partial pressure of helium, we use the formula:

$P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}$

Putting values in above equation, we get:

$745=245+119+163+p_{He}\\p_{He}=218mmHg$

Hence, the partial pressure of Helium is 218 mmHg.

For Part B:

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

$218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g$

Hence, the mass of helium gas is 0.504 g.

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