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3 years ago

For the following questions protons are blue and neutrons are red

Chemistry

1 answer:

faltersainse [42]3 years ago

6 0

I can’t see the picture where’s it at

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A student decreases the temperature of a 556 c m cubed balloon from 278 K to 231 K. Assuming constant pressure, what should the

a_sh-v [17]

Answer:

The new volume of the balloon be $462\ cm^3$ .

Explanation:

We have,

Initial volume is 556 cm ³

The temperature of the balloon decreases from 278 K to 231 K.

We need to find the new volume of the balloon if the pressure is constant.

The Charles law states that at constant pressure, the volume of gas is directly proportional to its temperature i.e.

$V\propto T$ , P is constant

$\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}$

$V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{556\times 231}{278}\\\\V_2=462\ cm^3$

So, the new volume of the balloon be $462\ cm^3$ .

3 0

3 years ago

The glow emitted by a substance exposed to external radiation is called

frosja888 [35]

Fluorescence
when a substance absorbs external radiation in any form and gives of a glow or luminescence it is considered as Fluorescence.

7 0

3 years ago

HELP ASAP!! WILL MARK BRAINLIEST !!

8_murik_8 [283]

Answer:

The question is incomplete.

${ \tt{\% \: composition = \frac{molar \: mass}{molecular \: mass} \times 100\%}} \\$

4 0

3 years ago

A solution contains 1.817 mg of CoSO4 (155.0 grams/mole) per mL. Calculate the volume (in mL) of 0.009795 M Zn2 needed to titrat

Nana76 [90]

Answer:

85.952 ml $Zn^2^+$ needed to titrate the excess complexing reagent .

Explanation:

Lets calculate

After addition of 80 ml of EDTA the solution becomes = 20 + 70 = 90 ml

As the number of moles of $CoSO_4$ = $\frac{Given mass }{molar mass}$

= $\frac{1.817}{155}$

=0.01172

Molarity = $\frac{no. of moles}{volume of solution}$

= $\frac{0.01172}{20}$

=0.000586 moles

Excess of EDTA = concentration of EDTA - concentration of CoSO4

= 0.009005 - 0.000586

= 0.008419 M

As M1V1 ( Excess of EDTA ) = M2V2 $(Zn^2^+)$

$0.008419\times100ml=0.009795\times V2$

$V2=\frac{0.008419\times100}{0.009795}$

V2 =85.952 ml

Therefore , 85.952 ml $Zn^2^+$ needed to titrate the excess complexing reagent .

3 0

3 years ago

A 50.0 mL sample of 1.54×10−2 M NaSO4 is added to 50.0 mL of 1.28×10−2 Ca(NO3)2. What percentage of the Ca2+ remains unprecipita

Gwar [14]

The reaction would be as shown below;
Na2SO4 + Ca(NO3)2 = CaSO4 + 2 NaNO3
The moles of NaSO4 will be;
= 0.05 × 0.0154 = 0.00077 moles
While the number of moles of Ca(NO3)2
= 0.05 × 0.0128 = 0.00064 moles
The mole ratio of sodium sulfate and calcium nitrate is 1:1
Ca(NO3)2 is the limiting reactant, so ignoring the Ksp of CaSO4, zero percent of the Ca^2+ ions remain unprecipitated.

3 0

3 years ago