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4 years ago

7

How far can a cough travel?

1 answer:

jolli1 [7]4 years ago

6 0

Traveling upwards of 200 mph or 320 km/h, and accelerating in a matter of seconds, germs from a cough or sneeze can travel a great distance very quickly.

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12. If the total pressure exerted by 3 gases is 45 atm and each individual gas has the

Bumek [7]

Answer:

Partial pressure for each of the three gases, in the mixture is 15 atm

Explanation:

Remember that the total pressure of a mixture, is the sum of partial pressures from the gases contained in the mixture.

Our total pressure = 45 atm

The 3 gases have the same pressure, so we can propose this equation:

3x = 45 atm

where x is the partial pressure for each of the three gases.

x = 45/3 → 15 atm

3 0

4 years ago

What is the diffence between solids , liquids and gases ?

vfiekz [6]

Answer:

solid : it has molecules tightly packed

liquid : it has molecules loosely packed

gas : it has molecules very far from each other

please mark as brainliest . please !

6 0

3 years ago

a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat

Nostrana [21]

Answer:- $\frac{382.69J}{0C}$ .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, $\Delta T$ for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

$q=mc\Delta T$

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

$q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})$

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

$\Delta T$ for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

$q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)$

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, $q=C.\Delta T$

$C=\frac{q}{\Delta T}$

$C=\frac{1044.746J}{2.73^0C}$

$C=\frac{382.69J}{0C}$

So, the heat capacity of calorimeter is $\frac{382.69J}{0C}$ .

4 0

3 years ago

A cylinder with a moveable piston contains 0.552 mol of gas and has a volume of 259 mL . Part A What will its volume be if an ad

Naddik [55]

Answer:

The new volume will be 367mL

Explanation:

Using PV = nRT

V1 = 259mL = 0.000259L

n1 = 0.552moles

At constant temperature and pressure, the value is

P * 0.000259 = 0.552 * RT ------equation 1

= 0.552 / 0.000259

= 2131.274

V2 = ?

n2 = 0.552 + 0.232

n2 = 0.784mole

Using ideal gas equation,

PV = nRT

P * V2 = 0.784 * RT ---------- equation 2

Combining equations 1 and 2 we have;

V2 = 0.784 / 2131.274

V2 = 0.000367L

V2 = 367mL

7 0

3 years ago

What is the percent by mass of water in MgSO4 • 7H2O? A. 51.1% B. 195% C. 56.0% D. 21.0%

lbvjy [14]

Answer : The correct option is, (A) 51.1%

Explanation :

Mass percent : It is defined as the mass of the given component present in the total mass of the compound.

Formula used :

$\text{Mass} \%H_2O=\frac{\text{Mass of }H_2O}{\text{Mass of }MgSO_4.7H_2O}\times 100$

First we have to calculate the mass of $H_2O$ and $MgSO_4.7H_2O$ .

Mass of $H_2O$ = 18 g/mole

Mass of $7H_2O$ = 7 × 18 g/mole = 126 g/mole

Mass of $MgSO_4.7H_2O$ = 246.47 g/mole

Now put all the given values in the above formula, we get the mass percent of $H_2O$ in $MgSO_4.7H_2O$ .

$\text{Mass} \%H_2O=\frac{126g/mole}{246.47g/mole}\times 100=51.1$

Therefore, the mass percent of $H_2O$ in $MgSO_4.7H_2O$ is, 51.1%

6 0

3 years ago