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3 years ago

An individual picks up a cup and drinks from it. This is an example of __________.

Chemistry

2 answers:

Lesechka [4]3 years ago

6 0

Answer:

Explanation:

The correct answer is: A) "A concentric contraction because muscles cause the elbow joint and arm to move"

Softa [21]3 years ago

6 0

Answer:

A. a concentric contraction because muscles cause the elbow joint and arm to move.

Explanation:

Contraction is the phenomenon in which the muscles of our body get activated by the tension produced. When the muscle of our body shortens, caused by the tension applied on our muscle is called concentric contraction. The force produced by the muscle as it shortens bring the cup near our body. There it is an example of concentric contraction.

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Help me please omg I don’t know

Anna35 [415]

Answer:

5 1 2 4and 3 this is correct way

8 0

2 years ago

Ammonium perchlorate NH4ClO4 is a powerful solid rocket fuel, used in the Space Shuttle boosters. It decomposes into nitrogen N2

patriot [66]

Answer: 0.055 moles of $O_2$ are produced by the reaction of 0.055 mol of ammonium perchlorate.

Explanation:

The balanced chemical reaction for decomposition of ammonium perchlorate is:

$2NH_4ClO_4\rightarrow N_2+Cl_2+2O_2+4H_2O$

According to stoichiometry :

2 moles of $NH_4ClO_4$ produce = 2 moles of $O_2$

Thus 0.055 moles of $NH_4ClO_4$ will produce = $\frac{2}{2}\times 0.055=0.055moles$ of $O_2$

Thus 0.055 moles of $O_2$ are produced by the reaction of 0.055mol of ammonium perchlorate.

7 0

3 years ago

Answers to all of these

Paul [167]

Answer:

1. Percent composition of Al = 13.423 %

Percent composition of Zn = 28.02 %

Percent composition of Cl = 30.6 %

Percent composition of O = 41.3 %

3. The empirical formula is C₅O₁₆

4. Molecular Formula= P₄O₆

Explanation:

Part first :

Data Given

Formula of the Molecule = Al₂ (CrO₄)₃

% of Al₂ = ?

> First of all find the atomic masses of each component in a molecule

For Al₂ (CrO₄)₃ atomic masses are given below

Al = 27 g/mol

Cr = 52 g/mol

O = 16 g/mol

> Then find the total masses of each component

2 atoms of Al = 27 g/mol x 2

= 54 g/mol

3 atoms of Cr = 52 g/mol x 3

= 156 g/mol

12 atoms of O = 16 g/mol x 12

= 192 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Al₂ (CrO₄)₃ = [27x2 + 52x3 + 16x12]

Molar Mass of Al₂ (CrO₄)₃ = 402

Now to find the mass percent of Al

Formula used to find the Mass percent of a component

Percent composition of Al = mass of Al in Molecula / molar mass of Al₂(CrO₄)₃ x 100%

Put the values

Percent composition of Al = 54 (g/mol) / 402 (g/mol) x 100%

Percent composition of Al = 13.423 %

_______________________________________

Part 2

Data Given

Formula of the Molecule = Zn(ClO₃)₂

% Zn = ?

% Cl = ?

% O = ?

> First of all find the atomic masses of each component in a molecule

For Zn(ClO₃)₂ atomic masses are given below

Zn = 65 g/mol

Cl = 35.5 g/mol

O = 16 g/mol

> Then find the total masses of each component

1 atoms of Zn= 65 g/mol x 1

= 65 g/mol

2 atoms of Cl = 35.5 g/mol x

= 71 g/mol

6 atoms of O = 16 g/mol x 6

= 96 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Zn(ClO₃)₂ = [65x1 + 35.5x2 + 16x6]

Molar Mass of Zn(ClO₃)₂ = 232g/mol

Now to find the mass percent of of each component one by one

1. Formula used to find the mass percent of Zn

Percent composition of Zn= mass of Zn in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Zn = 65(g/mol) / 232 (g/mol) x 100%

Percent composition of Zn = 28.02 %

-------------------

2. Formula used to find the mass percent of Cl

Percent composition of Cl = mass of Cl in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Cl = 71 (g/mol) / 232 (g/mol) x 100%

Percent composition of Cl = 30.6 %

---------------------

3. Formula used to find the mass percent of O

Percent composition of O = mass of O in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of O = 96 (g/mol) / 232 (g/mol) x 100%

Percent composition of O = 41.3 %

________________________________________

Part 3:

Data Given

Percentage of C = 27.3 %

Percentage of O = 72.7 %

Emperical Formula of the compound = ?

Solution:

So the compound has 27.3 % C and 72% O

First, find the mass of each of the elements in 100 g of the Compound.

C = 27.3 g

O = 72 g

Now find how many moles are there for each element in 100 g of compound

For this molar mass are required

That is

C = 12 g/mol

O = 16 g/mol

Formula Used

mole of C = mass of C / Molar mass of C

mole of C = 27.3 / 12 g/mol

mole of C = 2.275

Formula Used

mole of O = mass of O / Molar mass of O

mole of O = 72g / 16 g/mol

mole of O = 7.2

Divide each one by the smallest number of moles

C = 2.275 / 2.275

C = 1

O = 7.2 / 2.275

O = 3.2

Multiply the mole fraction to a number to get the whole number.

C = 1 x 5 = 5

O = 3.2 x 5 = 16

So, the empirical formula is C₅O₁₆

______________________________________

Part 4

Data Given

Percentage of P= 56.38 %

Percentage of O = 43.62%

Molar Mass = 219.9g

Molecular Formula of the compound = ?

Solution:

First, find the mass of each of the elements in 100 g of the Compound.

Mass of P= 56.38g

Mass of O = 43.62g

Now find how many moles are there for each element in 100 g of compound

find the moles in total compounds

Formula Used

mole of P = mass of / Molar mass of P

mole of P = 56.38 g / 31 g/mol

mole of P = 1.818

Formula Used

mole of O = mass of O / Molar mass of O

mole of O = 43. 62 / 16 g/mol

mole of O = 2.7262

Now

first find the Emperical formula

Divide each one by the smallest number of moles

P = 1.818 /1.818

P= 1

for oxygen

O = 2.7262 / 1.818

O = 1.5

Multiply the mole fraction to a number to get the whole number.

P = 1 x 2 = 2

O = 1.5 x 2 = 3

So, the empirical formula is P₂O₃

Now

Find molar mass of the empirical formula P₂O₃

2 (31) + 3 (16) = 62 + 48 = 110

Now find that how many empirical units are in a molecular unit.

(219.9 g/mol) / ( 110 g/mol) = empirical units per molecular unit

empirical units per molecular unit = 1.999 =2

A here we get two empirical units in a molecular unit,

So the molecular formula is:

2 (P₂O₃) = P₄O₆

7 0

3 years ago

if 45.0 ml of 1.50 M Ca(OH)2 are needed to neutralize 25.0 ml of HI of unknown concentration, what is the molarity of the HI?

ipn [44]

Answer:

M of HI = 5.4 M.

Explanation:

We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

(XMV) acid = (XMV) base.

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

(XMV) HI = (XMV) Ca(OH)₂.

For HI; X = 1, M = ??? M, V = 25.0 mL.

For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.

∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = 5.4 M.

6 0

3 years ago

How to test for sugar in your gingerbread

N76 [4]

Answer:

first of all we have to put sugar in bread and we have to add some ginger paste over it

4 0

3 years ago