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3 years ago

Why does a plant have a cell wall and a cell membrane?ASAP Pls have to turn it in at 11:00pm

Chemistry

1 answer:

Aleonysh [2.5K]3 years ago

4 0

Answer:

The cell wall surrounds the plasma membrane and provides the cell tensile strength and protection. ... The cell wall surrounds the plasma membrane of plant cells and provides tensile strength and protection against mechanical and osmotic stress.

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The coolest stars in the sky are ____ in color

Mekhanik [1.2K]

Red is the answer in the blank

4 0

3 years ago

Read 2 more answers

What is the mass of 8.23 x 10^23 atoms of Ag

Gnom [1K]

Answer:

$\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}$

Explanation:

Convert Atoms to Moles

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

$\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

$8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

$8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}$

$8.23 *10^{23} *\frac{1 \ mol \ Ag}{6.022*10^{23} }$

Condense into one fraction.

$\frac{8.23 *10^{23} }{6.022*10^{23} } \ mol \ Ag$

$1.366655596 \ mol \ Ag$

Convert Moles to Grams

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

Ag: 107.868 g/mol

Let's make another ratio using this information.

$\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

Multiply by the number of moles we calculated.

$1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

The moles of silver cancel out.

$1.366655596 *\frac {107.868 \ g \ Ag}{1 }$

$1.366655596 * {107.868 \ g \ Ag}$

$147.4184058 \ g\ Ag$

Round

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

147.4184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

$147 \ g\ Ag$

8.23 ×10²³ silver atoms are equal to approximately 147 grams.

3 0

3 years ago

A 20.0 mL solution of NaOH is neutralized with 24.1 mL of 0.200 M HBr. What is the concentration of the original NaOH solution

Alinara [238K]

Answer:

0.241 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

HBr + NaOH —> NaBr + H₂O

From the balanced equation above,

The mole ratio of acid, HBr (nₐ) = 1

The mole ratio of base, NaOH (n₆) = 1

Finally, we shall determine the concentration of the NaOH solution. This can be obtained as follow:

Volume of base, NaOH (V₆) = 20 mL

Volume of acid, HBr (Vₐ) = 24.1 mL

Concentration of acid, HBr (Cₐ) = 0.2 M

Concentration of base, NaOH (C₆) =?

CₐVₐ / C₆V₆ = nₐ/n₆

0.2 × 24.1 / C₆ × 20 = 1/1

4.82 / C₆ × 20 = 1

Cross multiply

C₆ × 20 = 4.82

Divide both side by 20

C₆ = 4.82 / 20

C₆ = 0.241 M

Therefore, the concentration of the NaOH solution is 0.241 M

8 0

3 years ago

Consider the following intermediate chemical equations.

QveST [7]

Answer: 250 kJ

Explanation: According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$P_4(s)+6Cl_2\rightarrow 4PCl_3$ $\Delta H_1=-2439kJ$ (1)