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2 years ago

I need help on science lol.

Chemistry

2 answers:

Anit [1.1K]2 years ago

6 0

Answer:

yes your right its is 4

Explanation:

Lena [83]2 years ago

5 0

Answer:

its 4

Explanation:

you had it right because if its at point 1 the other side has nothing so point 4 will go up

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What is the density of ice at -10°C?

Alisiya [41]

Answer:

3.98 C my friend you welcome

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3 years ago

Knowledge of atomic structure has evolved over time. Thomson discovered that atoms contain electrons, Rutherford discovered that

sergejj [24]

I believe the correct response would be A. The cumulative nature of science. How experimental evidence and ideas made by many scientists have accumulated and or built up to have a more understandable and or clear view of a particular scientific topic, and or principle.

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3 years ago

Read 2 more answers

gogolik [260]

Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

Explanation:

An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.

A Reductant thus exactly the opposite.

Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:

I2 --> 2I-.

The oxidation number reduced from 0 to -1.

In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.

Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.

7 0

3 years ago

HC2H3O2 (aq) + H2O (l) ⇔ C2H3O2- (aq) + H3O+ (aq) Ka = 1.8 x 10-5

marin [14]

The concentration of [H3O⁺]=2.86 x 10⁻⁶ M

<h3>Further explanation</h3>

In general, the weak acid ionization reaction

HA (aq) ---> H⁺ (aq) + A⁻ (aq)

Ka's value

$\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}$

Reaction

HC₂H₃O₂ (aq) + H₂O (l) ⇔ (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵

$\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}$

[H₃O⁺]=2.86 x 10⁻⁶ M

5 0

3 years ago

A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure

Lesechka [4]

Answer:

the final volume of the gas is $V_2$ = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure $P_{ext}$ :

$P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}$

$P_{ext} = 1.03 \ atm$

The workdone W = $P_{ext}$ V

The change in volume ΔV= $\dfrac{W}{P_{ext}}$

ΔV = $\dfrac{130.1 \ J \times \dfrac{1 \ L \ atm}{ 101.325 \ J} }{1.03 \ atm }$

ΔV = $\dfrac{1.28398717 }{1.03 }$

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial volume = 61.5 mL

The change in volume V is $\Delta V = V_2 -V_1$

$- V_2= - \Delta V -V_1$

multiply through by (-), we have:

$V_2= \Delta V+V_1$

$V_2$ = 1250 mL + 61.5 mL

$V_2$ = 1311.5 mL

∴ the final volume of the gas is $V_2$ = 1311.5 mL

5 0

2 years ago