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3 years ago

Which of these statements is not true?

Chemistry

1 answer:

STALIN [3.7K]3 years ago

4 0

Answer:

Explanation:

They can have different neutron numbers, not proton numbers. If they were to have different proton numbers it would be classified as a different element.

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In order to perform a chemical reaction, 225 mL of 0.500 M lead (II) nitrate is required. How much 5.00 M stock solution is requ

vladimir1956 [14]

Answer:

what

Explanation:

what ````

4 0

3 years ago

A mixture consists of 28% oxygen, 14% hydrogen, and 58% nitrogen by volume. A sample of this mixture has a pressure of 4.0 atm i

stepladder [879]

Answer:

C) 1.3 mol

Explanation:

Using gas law we can find the initial moles of the sample of the mixture, as follows:

PV = nRT

PV / RT = n

Where P is pressure: 4.0atm

V is volume: 9.6L

R is gas constant: 0.082atmL/molK

T is absolute temperature: 300K

And n are moles of the gas

PV / RT = n

4.0atm*9.6L / 0.082atmL/molK300K = n

n = 1.56moles of the mixture of the gas are present into the 9.6L container

Now, 14% of this gas is hydrogen that was removed of the system, that is:

1.56mol*14% = 0.22 moles of hydrogen are removed.

Thus, moles of gas that remains in the container are:

1.56mol - 0.22mol = 1.34mol.

Right answer is:

6 0

3 years ago

Read 2 more answers

How many moles of lithium nitrate (LINO3) are in 256 mL of a 0.855 M solution?

sweet [91]

Answer:

There are 0.219 mol of LINO3

7 0

3 years ago

3. How can you decrease the pressure of a gas in a container without changing the volume of the gas?

inna [77]

Answer:

reduce the temperature of the gas

Explanation:

when you reduce the temperature of the gas the pressure will decrease

8 0

3 years ago

A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene

muminat

Answer:

Molecular weight of the compound = 372.13 g/mol

Explanation:

Depression in freezing point is related with molality of the solution as:

$\Delta T_f = K_f \times m$

Where,

$\Delta T_f$ = Depression in freezing point

$K_f$ = Molal depression constant

m = Molality

$\Delta T_f = K_f \times m$

$1.33 = 5.12 \times m$

m = 0.26

Molality = $\frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}$

Mass of solvent (toluene) = 15.0 g = 0.015 kg

$0.26 = \frac{Mole\ of\ compound}{0.015}$

Moles of compound = 0.015 × 0.26 = 0.00389 mol

$Mol = \frac{Mass\ in\ g}{Molecular\ weight}$

Mass of the compound = 1.450 g

$Molecular\ weight = \frac{Mass\ in\ g}{Moles}$

Molecular weight = $\frac{1.450}{0.00389} = 372.13\ g/mol$

4 0

3 years ago