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2 years ago

Someone Help Me Plz. I'm not sure what the answer to this question is

Chemistry

1 answer:

ratelena [41]2 years ago

3 0

Answer:

I’m pretty sure the answer is c

Explanation:

Sorry if I’m wrong

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The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution

Alex_Xolod [135]

The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is 2.64 x 10⁻⁵ M.

4 0

3 years ago

Read 2 more answers

What is the empirical formula for a compound that is 31.9% potassium, 28.9% chlorine, and 39.2% oxygen?

notsponge [240]

The empirical formula for a compound is KClO3

Explanation
find the moles of each element
moles = % composition/molar mass

molar mass of of potassium =39g/mol ,chlorine = 35.5 g/mol, oxygen =16 g/mol

moles of potassium = 31.9 / 39 = 0.818 moles
moles of chlorine = 28.9/35.5 = 0.814 moles
moles of oxygen = 39.2/ 16 = 2.45 moles

find the mole ratio by dividing with the smallest mole = 0.814 moles

potassium = 0.818/0.814 =1
chlorine = 0.814/0.814 = 1
oxygen = 2.45 /0.814 =3

the empirical formula is therefore = KClO3

7 0

3 years ago

A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

OLEGan [10]

Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

Now, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$

Also,

mass of Ni, m = 15.0 g,

Initial temperature, $t_{i} = 100^{o}C$ ,

Final temperature, $t_{f}$ = ?

Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

$q_{water}(gain) = -q_{alloy}(lost)$

$55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$ = -( $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$ )

$t_{f} = \frac{25.9^{o}C}{1.029}$

= $25.2^{o}C$

Thus, we can conclude that the final temperature of nickel and water is $25.2^{o}C$ .

6 0

3 years ago

Cell membrane? (plz help)!!

VMariaS [17]

That’s ez that’s 1st grade math the answer is none

5 0

3 years ago

2NO + 3MnO2 + 4H â 2NO3- + 3Mn2 + 2H2O For the above redox reaction, assign oxidation numbers and use them to identify the eleme

mixer [17]

Answer:

Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).

Explanation:

Hello there!

In this case, according to the given redox reaction, we rewrite it as a convenient first step:

$2NO + 3MnO_2 + 4H^+ \rightarrow 2NO_3^- + 3Mn^{2+} + 2H_2O$

Next, we assign the oxidation numbers as follows:

$2N^{2+}O^{2-} + 3Mn^{4+}O^{-2}_2 + 4H^+ \rightarrow 2(N^{5+}O^{2-}_3)^- + 3Mn^{2+} + 2H^+_2O^{2-}$

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).

Regards!

4 0

3 years ago