Help anyone? :) please

3. For a boat to float on water, it must have a density lower than the density of water. Compare the densities of your unknown l

Fittoniya [83]

Answer: No

Explanation: if a boat was made of the unknown solid it would not float because it’s density was much higher than the density of water.

4 0

3 years ago

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What volume of .15M NaOH could be produced with only 5 grams of NaOH and water?

fenix001 [56]

The volume of 0.15M NaOH that could be produced with only 5 grams of NaOH and water is 0.83L.

<h3>How to calculate volume?</h3>

The volume of a substance can be calculated using the following expression:

Molarity = number of moles ÷ volume

The number of moles of NaOH can be calculated by dividing the mass of NaOH by its molar mass.

Molar mass of NaOH = 40g/mol

n = 5/40

n = 0.125moles

Volume of NaOH = 0.125mol ÷ 0.15M

Volume = 0.83L

Therefore, the volume of 0.15M NaOH that could be produced with only 5 grams of NaOH and water is 0.83L.

Learn more about volume at: brainly.com/question/905400

4 0

2 years ago

Suppose you have a solution that might contain any or all of the following cations: Cu 2, Ag , Ba 2, and Mn2 . The addition of H

SVEN [57.7K]

Answer:

Ag⁺, Ba²⁺,

Explanation:

We can solve this question using the solubility rules:

When an Halide as Br- is added to a solution, the ions that can be precipitate are Ag⁺, Hg₂²⁺ and Pb²⁺.

That means the first ion present is Ag⁺

When sulfates, SO₄²⁻ are added, the ions that precipitates are: Ag⁺, Ca²⁺, Sr²⁺, Ba²⁺, Hg₂²⁺ and Pb²⁺

The second ion present is Ba²⁺

Hydroxides of Cu²⁺ and Mn²⁺ are insolubles but as no precipitate are formed when the solution is strongly alkaline those ions are not present.

3 0

3 years ago

What is the maximum mass of PH3 that can be formed when 62.0g of phosphorus reacts with

Luda [366]

Answer: The mass of $PH_3$ produced is 45.22 g

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

For $P_4$ :

Given mass of $P_4$ = 62.0 g

Molar mass of $P_4$ = 124 g/mol

Putting values in equation 1, we get:

$\text{Moles of }P_4=\frac{62.0g}{124g/mol}=0.516mol$

For $H_2$ :

Given mass of $H_2$ = 4.00 g

Molar mass of $H_2$ = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2mol$

The chemical equation follows:

$P_4(g)+6H_2(g)\rightarrow 4PH_3(g)$

By stoichiometry of the reaction:

If 6 moles of hydrogen gas reacts with 1 mole of $P_4$

So, 2 moles of hydrogen gas will react with = $\frac{1}{6}\times 2=0.333mol$ of $P_4$

As the given amount of $P_4$ is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 6 moles of $H_2$ produces 4 mole of $PH_3$

So, 2 moles of $H_2$ will produce = $\frac{4}{6}\times 2=1.33mol$ of $PH_3$

We know, molar mass of $PH_3$ = 34 g/mol

Putting values in equation 1, we get:

$\text{Mass of }PH_3=(1.33mol\times 34g/mol)=45.22g$

Hence, the mass of $PH_3$ produced is 45.22 g

6 0

3 years ago

WILL MARK B!

melisa1 [442]

Answer:

Bro don't cheat in online exams bro

3 0

3 years ago