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1 year ago

How many molecules are in a 189 g sample of carbon tetrabromide, CBr4 ?

Chemistry

1 answer:

anyanavicka [17]1 year ago

5 0

Answer:

3.4 × 10²³ molecules of CBr₄

Explanation:

Given data:

Mass of CBr₄ = 189 g

Number of molecules = ?

Solution:

First of all we will calculate the number of moles.

Number of moles = mass / molar mass

Number of moles = 189 g/ 331.63 g/mol

Number of moles = 0.6 mol

Now the given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

Foe 0.6 moles of CBr₄:

0.6 mol × 6.022 × 10²³ molecules of CBr₄ / 1 mol

3.4 × 10²³ molecules of CBr₄

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As illustrated, the below manometer consists of a gas vessel and an open-ended U-tube containing a nonvolatile liquid with a den

STALIN [3.7K]

Answer:

1.01atm is the pressure of the gas

Explanation:

The difference in heights in the two sides is because of the difference in pressure of the enclosed gas and the atmospheric pressure. This difference is in mm of the nonvolatile liquid. The difference in mm Hg is:

32.3mm * (0.993g/mL / 13.6g/mL) = 2.36mmHg

As atmospheric pressure is 765mm Hg and assuming the gas has more pressure than the atmospheric pressure (There is no illustration), the pressure of the gas is:

765mm Hg + 2.36mm Hg = 767.36 mmHg

In atm:

767.36 mmHg * (1atm / 760 mmHg) =

1.01atm is the pressure of the gas

5 0

1 year ago

1. The specific heat capacity of iron is 0.461 J g–1 K–1 and that of titanium is 0.544 J g–1 K–1. A sample consisting of a mixtu

mezya [45]

Answer:

The answer is 80,1 °C

Explanation:

Let´s start from the mass of the sample and the heat capacities:

First of all, we must calculate an average heat capacity. That's because we have a mixture and it is unknown the heat capacity of the whole sample.

The way we should do this calculation is as follows:

(1) $H_{average}=Mass Fraction_{first component}* H_{first component}+MassFraction_{secondcomponent}*H_{second component}$

For example, the mass fraction of Fe is simply:

(2) $MassFraction_{Fe}=\frac{10g Fe}{10g Fe + 10gTi}=0.5$

If you combine the equations (1) and (2) you have:

(3) $H_{average}=0.5*0.461+0.5*0.544=0.5025\frac{J}{g-K}$

Once calculated the average heat capacity we can solve the problem taking into account the corresponding equation:

(4) $Q=m*H_{average}*(T_2-T_1)$

Remember that:

Q: Heat gained or lost

m: Mass of the sample you want to analize

$H_{average}$ : The value obtained in equation (3)

$T_2$ : Final temperature of the sample

$T_1$ : Initial temperature of the sample

Now we must replace the problem data in equation (4)

Take into account:

Heat gained in a system have a positive value
Heat lost in a system have a negative value

In this problem the sample loses 200 J, for this reason $Q=-200J$

The mass of the whole sample is: 10g of Fe + 10g of Ti = 20g of sample
The temperatures must be in absolute units of temperature (these are: rankine or kelvin)
The initial temperature of the system is 100°C or 373K

Now we are ready to use equation (4):

(5) $-200J=20g*0.5025\frac{J}{g*K} *(T_2-373K)$

It is clear that the unknown in equation (5) is $T_2$

The next step is to calculate $T_2$ . Don't forget the signs; these are important.

Key concept: Since the system is loosing heat, the final temperature of the system ( $T_2$ ) should be lower than the initial temperature ( $T_1$ )

7 0

2 years ago

Read 2 more answers

Balance the following redox reaction occurring in an acidic solution. The coefficient of Cr2O72−(aq) is given. Enter the coeffic

ollegr [7]

Answer:

Cr₂O₇²⁻ (aq) + 6Ti³⁺ (aq) + 2 H⁺(aq) → 2 Cr³⁺ (aq) + 6TiO²⁺(aq) + H2O(l)

Explanation:

Cr₂O₇²⁻ (aq) + Ti³⁺ (aq) + H⁺ (aq) → Cr³⁺ (aq) + TiO²⁺ (aq) + H₂O(l)

This is the reaction, without stoichiometry.

We have to notice the oxidation number of each element.

In dicromate, Cr acts with +6 and we have Cr³⁺, so oxidation number has decreased. .- REDUCTION

Ti³⁺ acts with +3, and in TiO²⁺ acts with +4 so oxidation number has increased. .- OXIDATION

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

To change 6+ to 3+, Cr had to lose 3 e⁻, but as we have two Cr, it has lost 6e⁻. As we have 7 Oxygens in reactant side, we have to add water, as the same amount of oxygens atoms we have, in products side. Finally, we have to add protons in reactant side, to ballance the H.

H₂O + Ti³⁺ → TiO²⁺ + 1e⁻ + 2H⁺

Titanium has to win 1 e⁻ to change 3+ to 4+. We had to add 1 water in reactant, and 2H⁺ in products, to get all the half reaction ballanced.

Now we have to ballance the electrons, so we can cancel them.

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) .1

(H₂O + Ti³⁺ → TiO²⁺ + 1e⁻ + 2H⁺) .6

Wre multiply, second half reaction .6

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O

6H₂O + 6Ti³⁺ → 6TiO²⁺ + 6e⁻ + 12H⁺

Now we can sum, the half reactions:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 6H₂O + 6Ti³⁺ → 6TiO²⁺ + 6e⁻ + 12H⁺ + 2Cr³⁺ + 7H₂O

Electrons are cancelled and we can also operate with water and protons

7H₂O - 6H₂O = H₂O

14 H⁺ - 12H⁺ = 2H⁺

The final ballanced equation is:

2H⁺ + Cr₂O₇²⁻ + 6Ti³⁺ → 6TiO²⁺ + 2Cr³⁺ + H₂O

8 0

2 years ago

If you are given the assignment to cut raw poultry for your entire shift, when should you sanitize the counter top, cutting boar

MrMuchimi

Sanitizing is essential in addition to cleaning poultry equipment. This is done only at the beginning and ending of your shift while cleaning should be done from time to time. Prior to sanitization, all surfaces must be clean and thoroughly rinsed to remove any detergent residue. Keep in mind that unclean surface cannot be sanitized. Sanitize refers to reduction of unwanted microorganisms to levels considered safe from a public health.

8 0

2 years ago

Read 2 more answers

A student mixed 20.00 grams of calcium nitrate, 10.00 grams of sodium nitrate, and 50.00 grams of aluminum nitrate in a 5.00 Lit

My name is Ann [436]

Answer:

$M=0.213M$

Explanation:

Hello,

In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:

$n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-$

$n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-$

$n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-$

We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:

$n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-$

Finally, we compute the molarity:

$M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M$

Regards.

6 0

2 years ago