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1 year ago

) A round or oval hole through a bone, which contains blood vessels and/or nerves, is called a __________.

1 answer:

8 0

The answer is foramen, which means opening.

This is term used in anatomy.

Its plural is foramina.

The foramina tipically allow the connection of two parts of the body through bloed vessels (arteries, veins) or nerves.

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Titanium dioxide, TiO₂, reacts with carbon and chlorine to give gaseous TiCl₄: TiO₂+2C+2CI₂−TiCI₄+2CO The reaction of 7.39 kg ti

Dima020 [189]

Answer:

17.57kg of $TiCl_{4}$ and its percentage yield is 81.0%

Explanation:

Through the reaction you can get the theoretical amount of $TiCl_{4}$ that must be produced.

$7.39kgTiO_{2}x\frac{1kmolTiO_{2} }{79.867kgTiO_{2}}x \frac{1kmolTiCl_{4}}{1kmolTiO_{2}}x\frac{189.867kgTiCl_{4} }{1kmolTiCl_{4}}=17.57kgTiCl_{4}$

If the amount obtained is less than the theoretical amount, it means that the initial sample was not 100% pure. Now the actual amount obtained is compared with the theoretical amount using a percentage

$yield=\frac{actual amount}{theoretical amount}x100= \frac{14.24kg}{17.57kg}x100$ =81.0%

3 0

1 year ago

James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o

gtnhenbr [62]

This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g

% O = 48/84 * 100 = 57.14%

5 0

2 years ago

Read 2 more answers

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

5 0

1 year ago

A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr

s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

$P=p_{N_2}+p_{O_2}+p_{Ar}$

$p_{N_2}=P\times \chi_{N_2}$

$p_{O_2}=P\times \chi_{O_2}$

$p_{Ar}=P\times \chi_{Ar}$

where,

$P$ = total pressure = 3.9 bar

$p_{N_2}$ = partial pressure of nitrogen gas

$p_{O_2}$ = partial pressure of oxygen gas

$p_{Ar}$ = partial pressure of argon gases

$\chi_{N_2}$ = Mole fraction of nitrogen gas = 0.25

$\chi_{O_2}$ = Mole fraction of oxygen gas = 0.65

$\chi_{Ar}$ = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

$p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar$

Partial pressure of oxygen gas :

$p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar$

Partial pressure of argon gas :

$p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar$

7 0

1 year ago

Rank the following compounds in order of reactivity toward nitration with HNO3:_______. thiophene, benzene, 3-methylthiophene, a

siniylev [52]

Answer:

3-methylthiophene > thiophene > benzene > 2-methylfuran

Explanation:

Primarily, five membered heterocyclic aromatic rings undergo nitration at carbon-2. This is because, nitration at carbon-2 leads to the formation of three resonance structures while attack at carbon-3 yields only two resonance structures, hence it is less stabilized.

The presence of a methyl group which donates electrons promotes the stabilization of the cation formed in the nitration of 3-methylthiophene.

2-methylfuran is the least reactive towards nitration because the 2-position has been blocked by a methyl group.

3 0

1 year ago