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3 years ago

You have a square with a Perimeter of 16 in,, what is the Area of the square after scaling it by 5?

Chemistry

1 answer:

Novosadov [1.4K]3 years ago

8 0

Answer:

f the figure is a square with a perimeter of 16 inches, then each side of the square is 4 inchest in length. Area is found by multiplying the length x the width. For a square, the length and width are the same. A = 16 sq

Explanation:

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Give the spectator ions for the reaction that occurs when aqueous solutions of H 2SO 4 and KOH are mixed.

Sever21 [200]

Answer:

The spectator ions is: $K^+$ and $SO^{2-}_4$

Explanation:

The equation of reaction between H₂ SO₄ and KOH is:

$H_2SO_{4(aq)} + 2KOH _{aq} \to K_2SO_{4(aq)} +2H_2O _{(l)}$

Rewriting this equation as ionic;

$[2H^{+} + SO^{2-}_4 + 2K^+ +2OH^- \to 2K^+ SO_4^{2-} + 2H_2O ]$

Spectators ions are ions present on both sides of the ionic equation by the same quantity but do not take part in the net reaction.

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3 years ago

An atom with a full outer shell will be _______ than an atom with an incomplete outer shell.

skelet666 [1.2K]

B) They are less reactive

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3 years ago

Which structures join with the cell’s membrane during exocytosis?

erma4kov [3.2K]

Answer:Exocytosis is also used to integrate new proteins into the cell membrane. In this process, the new protein is formed inside the cell, and migrates to phospholipid bilayer of the vesicle. The vesicle, containing the new protein as a part of the phospholipid bilayer, fuses with the cell membrane.

Explanation:

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3 years ago

How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group

OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

P= 1.00 atm
V= 150 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 $\frac{atm*L}{mol*K}$ *273 K

Solving:

$n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}$

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

$mass=\frac{6.69 moles*70.9 grams}{1 mole}$

mass= 474.321 grams ≅ 474 grams

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

4 0

3 years ago

What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a

Vadim26 [7]

Answer:

$\boxed {\boxed {\sf 333 \ grams}}$

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

$Q= mc \Delta T$

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

ΔT= final temperature - initial temperature

The sample was heated from 58.8 degrees Celsius to 88.9 degrees Celsius.

ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

Q= 4500.0 J
c= 0.4494 J/g°C
ΔT = 30.1 °C

Substitute these values into the formula.

$4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)$

Multiply on the right side of the equation. The units of degrees Celsius cancel.

$4500.0 \ J = m (13.52694 J/g)$

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

$\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}$

The units of Joules cancel.

$\frac {4500.0 \ J }{13.52694 J/g}= m$

$332.6694729 \ g =m$

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

$333 \ g \approx m$

The mass of the sample of metal is approximately 333 grams.

8 0

2 years ago

You have a square with a Perimeter of 16 in,, what is the Area of the square after scaling it by 5?​

You have a square with a Perimeter of 16 in,, what is the Area of the square after scaling it by 5?