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2 years ago

Contractile proteins are found in _____.

Chemistry

2 answers:

Snowcat [4.5K]2 years ago

7 0

Answer: In The Muscles

FinnZ [79.3K]2 years ago

5 0

C. They are found in muscles.

You might be interested in

Convert 2.0 M of Phenobarbital sodium (MW: 254 g/mole) solution in water into % w/v and ratio strengths.

Travka [436]

Answer:

The concentration is 50,8 % w/v and radio strengths = 1,96.

Explanation:

Phenobarbital sodium is a medication that could treat insomnia, for example.

2,0 M of Phenobarbital sodium means 2 moles in 1L.

The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means 1g in r mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:

2 moles × $\frac{254 g}{1 mole}$ = 508 g of Phenobarbital sodium.

1 L × $\frac{1000 mL}{ 1 L}$ = 1000 mL of solution

Thus, % w/v is:

$\frac{508 g}{1000 mL}$ × 100 = 50,8 % w/v

And radio strengths:

$\frac{1000 mL}{508 g}$ = 1,96. Thus, you have 1 g in 1,96 mL

I hope it helps!

5 0

2 years ago

How many acidic protons are there in 0.6137 g of KHP?

andreyandreev [35.5K]

0.6137 g of KHP contains 1.086 × 10^21 acidic protons.

Number of moles of KHP = mass of KHP/molar mass of KHP

Molar mass of KHP = 204.22 g/mol

Mass of KHP = 0.6137 g

Number of moles of KHP = 0.6137 g/204.22 g/mol = 0.003 moles of KHP

Now, 1 each molecule of KHP contains 1 acidic proton.

For 0.003 moles of KHP there are; 0.003 × 1 × NA

Where NA is Avogadro's number.

So; 0.003 moles of KHP contains 0.003 × 1 × 6.02 × 10^23

= 1.086 × 10^21 acidic protons.

Learn more: brainly.com/question/16672114

7 0

2 years ago

Determine net ionic equations, if any, occuring when aqueous solutions of the following reactants are mixed. Select "True" or "F

nirvana33 [79]

Answer:

For 1: The correct answer is False.

For 2: The correct answer is True.

For 3: The correct answer is True.

For 4: The correct answer is False.

For 5: The correct answer is True.

Explanation:

Net ionic equation of any reaction does not include any spectator ions. If no net ionic equation is formed, it is said that no reaction has occurred.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form. Solids, liquids and gases do not exist as ions.

For 1: Lead(II) nitrate and sodium chloride

The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:

$Pb(NO_3)_2(aq.)+2NaCl(aq.)\rightarrow PbCl_2(s)+2NaNO_3(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+2Na^+(aq.)+2NO_3^-(aq.)$

As, sodium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)$

Hence, the correct answer is False.

For 2: Sodium bromide and hydrochloric acid

The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:

$NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)$

Ionic form of the above equation follows:

$Na^{+}(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)$

There are no spectator ions in the equation. So, the above reaction is the net ionic equation.

Hence, the correct answer is True.

For 3: Nickel (II) chloride and lead(II) nitrate

The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:

$Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)$

Ionic form of the above equation follows:

$Pb^{2+}(aq.)+2NO_3^-(aq.)+Ni^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+Ni^{2+}(aq.)+2NO_3^-(aq.)$

As, nickel and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+Cl^-(aq.)\rightarrow PbCl_2(s)$

Hence, the correct answer is True.

For 4: Magnesium chloride and sodium hydroxide

The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:

$MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)$

Ionic form of the above equation follows:

$Mg^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)$

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Mg^{2+}(aq.)+OH^-(aq.)\rightarrow Mg(OH)_2(s)$

Hence, the correct answer is False.

For 5: Ammonium sulfate and barium nitrate

The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:

$(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)$

Ionic form of the above equation follows:

$2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ba^{2+}(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)$

Hence, the correct answer is True.

3 0

2 years ago

A sample of a certain barium chloride hydrate, BaCl2.nH2O, has a mass of 7.62 g. When this sample is heated in a crucible, it is

Ratling [72]

Answer:

BaCl2.2H2O

Explanation:

Data obtained from the question include:

Mass of barium chloride hydrate (BaCl2.nH2O) = 7.62g

Mass of anhydrous barium chloride (BaCl2) = 6.48g

Next, we shall determine the mass of water in the barium chloride hydrate (BaCl2.nH2O). This can be obtained as follow:

Mass of water (H2O) = Mass of barium chloride hydrate (BaCl2.nH2O) – Mass of anhydrous barium chloride (BaCl2)

Mass of water (H2O) = 7.62 – 6.48

Mass of water (H2O) = 1.14g

Next, we shall write the balanced equation for the reaction. This is given below:

BaCl2.nH2O —> BaCl2 + nH2O

Next, we shall determine the masses of BaCl2.nH2O that decomposed and the mass H2O produced from the balanced equation.

Molar mass of BaCl2.nH2O = 137 + (35.5x2) + n[(2x1) + 16]

= 137 + 71 + n[2 + 16]

= (208 + 18n) g/mol

Mass of BaCl2.nH2O from the balanced equation = 1 x (208 + 18n) = (208 + 18n) g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = n x 18 = 18n g

Summary:

From the balanced equation above,

(208 + 18n) g of BaCl2.nH2O decomposed to produce 18n g of H2O.

Finally, we shall determine the formula for the hydrated compound as follow:

From the balanced equation above,

(208 + 18n) g of BaCl2.nH2O decomposed to produce 18n g of H2O.

But 7.62g of BaCl2.nH2O decomposed to produce 1.14g of H2O i.e

18n/(208 + 18n) = 1.14/7.62

Cross multiply

18n x 7.62 = 1.14(208 + 18n)

137.16n = 237.12 + 20.52n

Collect like terms

137.16n – 20.52n = 237.12

116.64n = 237.12

Divide both side by the coefficient of n i.e 116.64

n = 237.12/116.64

n = 2

Therefore the formula for the hydrate, BaCl2.nH2O is BaCl2.xH2O.

3 0

2 years ago

Calculate the mass in grams of 1.32x10^20 uranium atoms

scoundrel [369]

Uranium has an atomic weight of 238.03 g/mol. We have 1.32x10^20 atoms of uranium. We must convert this to moles by dividing Avogadro's number 6.022x10^23 atoms/mol. (1.32x10^20atoms)/(6.022x10^23atoms/mol) 2.19x10^-4moles of Uranium Now multiply this by the atomic weight of Uranium. 2.19x10^-4mol*238.03g/mol Grams of Uranium = .0522g

6 0

2 years ago