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lawyer [7]
3 years ago
9

PLEASEEEE HELP MEE!

Chemistry
1 answer:
Ugo [173]3 years ago
7 0

Answer:

2nd option

Explanation:

You might be interested in
a compound has 15.39 g of gold for every 2.77 g of chlorine. simplified there is _____ g of gold for every 1 g of chlorine
iren [92.7K]

Answer:

There is 5.56 g of gold for every 1 g of chlorine

Explanation:

The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction \frac{a}{b}

You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:

\frac{15.39 g  of gold}{2.77 g of chlorine}

The proportion is the equal relationship that exists between two reasons and is represented by:    \frac{a}{b}=\frac{c}{d}

This reads a is a b as c is a d.

To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:

\frac{15.39 g  of gold}{2.77 g of chlorine}=\frac{mass of gold}{1 g of chlorine}

Solving for the mass of gold gives:

mass of gold=1 g of chlorine*\frac{15.39 g  of gold}{2.77 g of chlorine}

mass of gold= 5.56 grams

So, <u><em>there is 5.56 g of gold for every 1 g of chlorine</em></u>

5 0
3 years ago
Sodium thiosulfate, Na 2S 2O 3, is used as a "fixer" in black and white photography. Identify the reducing agent in the reaction
Tatiana [17]

Answer: Reducing agent in the given reaction is S_{2}O^{2-}_{3}.

Explanation:

A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.

In the given reaction, oxidation state of sulfur in S_{2}O^{2-}_{3} is +2 and I_{2}(aq) has 0 oxidation state.

In S_{4}O^{2-}_{6}(aq) oxidation state of S is 2.5 and in 2I^{-}(aq) oxidation state of I is -1.

Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.

Thus, we can conclude that reducing agent in the given reaction is S_{2}O^{2-}_{3}.

8 0
3 years ago
What led Thomson to decide that Dalton's atomic model needed to be revised?
jeka94

Answer:

Discovery of electron while studying the properties of cathode ray by Thomson suggested that Dalton atomic model should be revised.

Explanation:

Electron was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

He constructed the glass tube and create vacuum in it. He applied electric current between electrodes. He noticed that a ray of particles coming from cathode to wards positively charged anode. This ray was cathode ray.

Properties of cathode ray:

The ray is travel in straight line.

The cathode ray is independent of composition of cathode.

When electric field is applied cathode ray is deflected towards the positively charged plate.

Hence it was consist of negatively charged particles.

Symbol= e-

Mass= 9.10938356×10-31 Kg

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

4 0
3 years ago
The combustion of acetylene gas is represented by this equation: 2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
MAVERICK [17]

Answer:

Approximately 2.46\; \rm mol.

Explanation:

Make use of the molar mass data (M({\rm C_2H_2}) = 26.04\; \rm g \cdot mol^{-1}) to calculate the number of moles of molecules in that 64.0\; \rm g of \rm C_2H_2:

\begin{aligned}n({\rm C_2H_2}) &= \frac{m({\rm C_2H_2})}{M} \\ &= \frac{64.0\; \rm g}{26.04\; \rm g\cdot mol^{-1}}\approx 2.46\; \rm mol\end{aligned}.

Make sure that the equation for this reaction is balanced.

Coefficient of \rm C_2H_2 in this equation: 2.

Coefficient of \rm H_2O in this equation: 2.

In other words, for every two moles of \rm C_2H_2 that this reaction consumes, two moles of \rm H_2O would be produced.

Equivalently, for every mole of \rm C_2H_2 that this reaction consumes, one mole of \rm H_2O would be produced.

Hence the ratio: \displaystyle \frac{n({\rm H_2O})}{n({\rm C_2H_2})} = \frac{2}{2} = 1.

Apply this ratio to find the number of moles of \rm H_2O that this reaction would have produced:

\begin{aligned}n({\rm H_2O}) &= n({\rm C_2H_2}) \cdot \frac{n({\rm H_2O})}{n({\rm C_2H_2})} \\ &\approx 2.46\; \rm mol \times 1 = 2.46\; \rm mol\end{aligned}.

3 0
3 years ago
The volume of a gas is reduced from 4 L to 0.5 L while the temperature is held constant. How does the gas pressure change?
liberstina [14]

Answer:

It increases by a factor of eight

Explanation:

When temperature is held constant, gas pressure changes according the volume, in undirectly proportion.

Volume increases →  Pressure decreases

Volume decreases → Pressure increases

As volume gas, was reducted from 4L to 0.5L, it was reduced by 1/8, so the pressure gas was increased by a factor of eight.

5 0
3 years ago
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