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Svetllana [295]
2 years ago
13

Answer truthfully:))​

Chemistry
2 answers:
salantis [7]2 years ago
6 0

Answer:

The first guy is correct. That's what I got too

Elina [12.6K]2 years ago
3 0

Answer:

<u>Formula</u><u>:</u> Velocity \:  V = f \lambda \\  Solution: = 5 \times 0.8 \\  = 4 \:  {ms}^{ - 1}

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ObIel WiLll unt COl.. USSMS A certain chemical reaction releases 31.2 kJ/g of heat for each gram of reactant consumed. How can y
aleksandr82 [10.1K]

Answer:

The expression to calculate the mass of the reactant is m = \frac{1.080kJ}{31.2kJ/g}

Explanation:

<em>The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant.</em> To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:

1,080J.\frac{1kJ}{10^{3}J } =1.080kJ

Then,

1.080kJ=31.2\frac{kJ}{g} .m\\m = \frac{1.080kJ}{31.2kJ/g}

5 0
3 years ago
Read 2 more answers
From the formula MgO how do you know that Mg is the metal?
Damm [24]

In binary ionic compounds the name of the cation (Metal) is first, so that’s how you know.

5 0
3 years ago
For the following electrochemical reaction: Al3+(aq) + 3e -&gt; Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -&gt; 2F (aq) Calculate
makvit [3.9K]

<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=2.87-(-1.66)=4.53V

Hence, the standard electrode potential of the cell is 4.53 V.

6 0
2 years ago
Which statement correctly describes the location and charge of the neutrons in an atom?
rodikova [14]

Answer:

o

Explanation:

6 0
3 years ago
After decaying for 48 hours, one-sixteenth (1/16) of the original mass of a radioisotope sample remains unchanged. What is the h
Hunter-Best [27]

The half-life of this radioisotope : 12 hr

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually radioactive elements have an unstable atomic nucleus.  

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

t=48 hr

\tt \dfrac{Nt}{No}=\dfrac{1}{16}

The half-life :

\tt \dfrac{1}{16}=\dfrac{1}{2}^{(48/t\frac{1}{2} )}\\\\(\dfrac{1}{2})^4=(\dfrac{1}{2})^{48/t\frac{1}{2}}\\\\4=48/t\frac{1}{2}\\\\t\frac{1}{2}=12~hr

7 0
2 years ago
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