Which organism is a producer? A:owl B:mouse C:grass D:snake please somebody help me

2. How many moles are in 8.30 x 1023 molecules of H2O?

daser333 [38]

Answer:

<h2>1.38 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

$n = \frac{8.30 \times {10}^{23} }{6.02 \times {10}^{23} } = \frac{8.30}{6.02} \\ = 1.3787 37...$

We have the final answer as

<h3>1.38 moles</h3>

Hope this helps you

8 0

3 years ago

How can a scientist make sure that the results are not a mistake in an experiment?

olga55 [171]

Repeat trials multiple times

8 0

3 years ago

g Given that 50.0 mL of 0.100 M magnesium bromide reacts with 13.9 mL of silver nitrate solution according to the unbalanced equ

ipn [44]

Answer:

0.719M AgNO₃

Explanation:

Based on the reaction:

MgBr₂ + 2AgNO₃ ⇄ 2AgBr + Mg(NO₃)₂

<em>1 mole of magnesium bromide reacts completely with 2 moles of AgNO₃</em>

<em />

To find molarity of AgNO₃ solution we need to determine moles of AgNO₃ and, as molarity is the ratio of moles over liter (13.9mL = 0.0139L). Now, to determine moles of AgNO₃ we need to use the reaction, thus:

<em>Moles AgNO₃:</em>

<em />

Moles of MgBr₂ are:

50.0mL = 0.050L * (0.100mol / L) = 0.00500 moles of MgBr₂.

As the silver nitrate reacts completely and 2 moles of AgNO₃ reacts per mole of MgBr₂:

0.00500 moles MgBr₂ * (2 moles AgNO₃ / 1 mole MgBr₂) =

0.0100 moles of AgNO₃ are in the solution.

And molarity is:

0.0100 moles AgNO₃ / 0.0139L =

3 0

3 years ago

Determine the empirical formula for a compound that is found to contain

Sphinxa [80]

Answer:

The empirical formula is PCl3

Explanation:

Mass of P is 30.97 g, thus 1.523 g of P equivalent to 0.05 moles of P

Mass of Cl is 35.45 g, thus 5.228 g of Cl equivalent to 0.15 moles of Cl

Therefore moles of P : moles of Cl = 0.05:0.15 = 1:3

Therefore the empirical formula, PCl3

7 0

3 years ago

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin

LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine $(C_3H_7NO_2)$ are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is $5.9^oC$ lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is $7.2^oC$ lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

<u>Answer:</u> The van't Hoff factor for potassium bromide in X is 1.63

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\Delta T_f=i\times K_f\times m$