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3 years ago

I don’t get it plz help

Chemistry

1 answer:

d1i1m1o1n [39]3 years ago

3 0

D=m/v. (Density equals mass over volume).
So in your case density = 4.2/6
Which equals 0.7
Your units will be grams over milliliters

So your answer is 0.7 g/ml

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Which of the following is true of atoms with very low electronegativity

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Answer:

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1 Electronegativity is the ability of an atom to attract electrons.

Explanation:

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2 years ago

A 0.250 g sample of hydrocarbon (containing only carbon and hydrogen) undergoes complete combustion to produce 0.845 g of CO2 an

melomori [17]

Answer:

Explanation:

We have to obtain the mass of carbon and hydrogen in CO2 and H2O respectively.

For carbon in CO2;

0.845 * 12/44 = 0.23 g

For hydrogen in H20;

0.173 * 2/18 = 0.019 g

We convert the masses to moles of carbon and hydrogen

For carbon - 0.23/ 12 = 0.019 moles

For hydrogen - 0.019/1 = 0.019 moles

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3 years ago

A sample of g of pure aluminum metal is added to mL of M hydrochloric acid. The volume of hydrogen gas produced at standard temp

MArishka [77]

Answer:

V = 11.2L are produced

Explanation:

... Sample of 27g of pure aluminium, 3added to 333 mL of 3.0 M HCl..

Based on the chemical reaction:

2Al(s) + 6HCl(aq) → 2AlC₃(aq) + 3H₂(g)

Where 3 moles of hydrogen are produced when 6 moles of hydrochloric acid reacts with 2 moles of Al.

To solve this question, we need to determine limiting reactant converting each reactant to moles. With limiting reactant and the chemical reaction we can find moles of hydrogen and its volume at STP (T=273.15K; P=1atm), thus:

Moles Al-Molar mass: 26.98g/mol-:

27g * (1mol / 26.98g) = 1mol of Al

Moles HCl:

333mL = 0.333L * (3mol/L) = 1mol HCl

For a complete reaction of 1 mole of HCl are required:

1mol HCl * (2mol Al / 6mol HCl) = 0.333 moles of Al. As there is 1 mole of Al, Al is in excess and HCl is limiting reactant.

Moles of Hydrogen produced are:

1mol HCl * (3 moles H₂ / 6 mol HCl) = 0.5moles H₂ are produced.

Using ideal gas law:

PV = nRT

V = nRT/P

Where V is volume

n are moles: 0.5mol

R is gas constant: 0.082atmL/molK

T is absolute temperature: 273.15K

P is pressure: 1atm.

Solving for V:

V = 0.5mol*0.082atmL/molK*273.15K / 1atm

<h3>V = 11.2L are produced</h3>

3 0

3 years ago

A sample of a gas in a rigid container has an initial pressure of 1.049 kPa and an initial temperature of 7.39 K. The temperatur

Doss [256]

Answer:- 4.36 kPa

Solution:- At constant volume, the pressure of the gas is directly proportional to the kelvin temperature.

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Where the subscripts 1 and 2 are representing initial and final quantities.

From given data:

$P_1$ = 1.049 kPa

$P_2$ = ?

$T_1$ = 7.39 K

$T_2$ = 30.70 K

For final pressure, the equation could also be rearranged as:

$P_2=\frac{P_1T_2}{T_1}$

Let's plug in the values in it:

$P_2=\frac{1.049kPa(30.70K)}{7.39K}$

$P_2$ = 4.36 kPa

So, the new pressure of the gas is 4.36 kPa.

5 0

3 years ago

Find the volume of a gas STP if it’s volume is 80.0 mL at 109 kpa and-12.5c

photoshop1234 [79]

Answer:

Approximately $8.38 \times 10^8\; \rm mL$ , which is the same as $8.38 \times 10^5 \; \rm L$ . Assumption: the behavior of this gas is ideal.

Explanation:

Initial state of the gas:

$T_\text{initial} = -12.5\; \rm ^\circ C = (-12.5 + 273.15)\; \rm K = 260.65\; \rm K$ .
$P_\text{initial} = \; \rm 10^9\; \rm kPa = 10^{12}\; \rm Pa$ .

STP state:

$T_\text{STP} = 0\; \rm ^\circ C = 273.15\; \rm K$ .
$P_\text{STP} = 10^5\; \rm Pa$ .

The initial state of this gas can be changed to STP state in two steps:

First, reduce the pressure from $10^{12}\; \rm Pa$ to $10^5\; \rm Pa$ .
Second, increase the temperature from $260.65\; \rm K$ to $273.15\; \rm K$ .

Assume that the gas acts like an ideal gas at all time. Also, assume that the number of gas particles did not change.

Assume that temperature stays the same when the pressure changes from $10^{12}\; \rm Pa$ to $10^5\; \rm Pa$ . By Boyle's Law, volume is inversely proportional to pressure when all other factors stay the same. In other words,

$\begin{aligned}V_\text{intermediate} &= V_\text{initial} \cdot \displaystyle \frac{P_{\text{initial}}}{P_{\text{STP}}} \\ &= 80.0\; \rm mL \times \frac{10^{12}\; \rm Pa}{10^5\; \rm Pa} = 8.00 \times 10^8\; \rm mL\end{aligned}$ .

After that, assume that pressure stays the same when the temperature changes from $\rm 260.65\; \rm K$ to $\rm 273.15\; \rm K$ . By Charles's Law, volume is proportional to pressure when all other factors stay the same. In other words,

$\begin{aligned}V_\text{STP} &= V_\text{intermediate} \cdot \displaystyle \frac{T_{\text{STP}}}{T_{\text{initial}}} \\ &= 8.00 \times 10^8\; \rm mL \times \frac{273.15\; \rm K}{260.65\; \rm K} \\ &\approx 8.38\times 10^8\; \rm mL = 8.38 \times 10^5\; \rm L\end{aligned}$ .

7 0

3 years ago