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3 years ago

PLZ Help asap!!!!

Chemistry

1 answer:

Naily [24]3 years ago

5 0

Answer:

C. 9.70

Explanation:

Hello,

This problem is a simple dilution problem which utilizes the formula M1V1= M2V2 (where M is molarity, V is volume). I have attached a helpful image as well.

M1 V1 = M2 V2

(18) (x) = (0.35) (500)

x = 9.72mL which is close enough to option C.

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What is the overall charge of an ion Pat has 12 protons 10 electrons and 14 neutrons

Sav [38]

Answer:

The over all charge on atom will be +2.

Explanation:

An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.

For example

X is the element having 12 protons 10 electrons and 14 neutrons.

The number of protons and electrons are not equal which means two electrons are lose by the given atom and it is present in the form of cation.

The over all charge on atom will be +2.

4 0

3 years ago

In the equation CH4 + 2O2 --> 2H2O + CO2 What is the mass of CO2 produced when 35g of O2 reacts?

Anestetic [448]

Answer:

24.06 g of CO₂

Explanation:

The balanced equation for the reaction is given below:

CH₄ + 2O₂ —> 2H₂O + CO₂

Next, we shall determine the mass of O₂ that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂.

Finally, we shall determine the mass of CO₂ produced by the reaction of 35 g of O₂. This can be obtained as follow:

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂.

Therefore, 35 g of O₂ will react to produce = (35 × 44)/64 = 24.06 g of CO₂.

Thus, 24.06 g of CO₂ were produced from the reaction.

8 0

2 years ago

Can someone please help

Ratling [72]

Answer:

1. c) shiny

2) True. Reactivity is a chemical property.

6 0

3 years ago

Read 2 more answers

5. How much heat is evolved when 24.9 g of propanol (C3H7OH) is burned? ΔHcomb = −2010 kJ/mol

3241004551 [841]

Answer:

At the global scale, they are a significant source of emitted carbon, contributing to global warming which could lead to biodiversity changes. ... The consequence of repeated burns is detrimental because it is a key factor in the impoverishment of biodiversity in rain forest ecosystems.

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3 years ago

Methyl salicylate is a common active ingredient in liniments such as ben-gay. It is also known as oil of wintergreen. It is made

arlik [135]

Answer: The empirical formula is $C_3H_3O$ .

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 12.24 g

Mass of $H_2O$ = 2.505 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, = $\frac{12}{44}\times 12.24=3.338g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, = $\frac{2}{18}\times 2.505=0.278g$ of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.278}{0.104}=3$

For H = $\frac{0.278}{0.104}=3$

For O = $\frac{0.104}{0.104}=1$

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is $C_3H_3O$ .

5 0

3 years ago