All categories

3 years ago

What are three elements that are in the same group

Physics

1 answer:

cluponka [151]3 years ago

3 0

Iodine, chloride, and bromide

You might be interested in

The percentage of fe in an iron ii sample was determined by redox titration with cr2o7 calculate the molarity of the standard k2

Archy [21]

The molarity is 0.018M and the percentage is 8.46%.

Net ionic reaction is

$\mathrm{Cr} 2 \mathrm{O}^{2-}+6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+} \rightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H} 2 \mathrm{O}$

$1 \mathrm{~mol}$ of $\mathrm{Cr} 2 \mathrm{O} 7(-2)$ reacts with $6 \mathrm{~mol}$ of $\mathrm{Fe}(2+)$ to form $6 \mathrm{~mol}$ of $\mathrm{Fe}(3+)$ and $2 \mathrm{~mol}$ of $\mathrm{Cr}(3+)$

Find molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$

Molarity = moles of $\mathrm{K} 2 \mathrm{Cr} 207 /$ Volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7(\mathrm{~L}$ )

Moles of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ = grams of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 /$ molar mass of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$

$=1.2275 \mathrm{~g} / 294.19 \mathrm{~mol}=0.0042 \mathrm{~g} / \mathrm{mol}$

Molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7=$ moles of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ / Volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7$ (L)

$=0.0042 \mathrm{~g} / \mathrm{mol} / 0.250 \mathrm{~L}=0.017 \mathrm{M}$

Find molarity of $\mathrm{Fe}$ (II)

Molarity of $\mathrm{Fe}(\mathrm{II})=6 \times$ molarity of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 \times$ volume of $\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O} 7 /$ volume of $\mathrm{Fe}$ (II)

$=6 \times 0.017 \mathrm{M} \times 0.03598 \mathrm{~L} / 0.200 \mathrm{~L}$

$=0.018 \mathrm{M}$

Find moles of $\mathrm{Fe}(\mathrm{II})$

Moles of $\mathrm{Fe}($ II) =$ molarity of $\mathrm{Fe}$ (II) $\times$ volume of $\mathrm{Fe}$ (II)

$=0.018 \mathrm{M} \times 0.200 \mathrm{~L}=0.004 \mathrm{~mol}$

Find mass of Fe(II)

Mass of $\mathrm{Fe}( II )=$ moles of $\mathrm{Fe}( II$ ) $\times$ molar mass of $\mathrm{Fe}(\mathrm{II})$

$=0.004 \mathrm{~mol} \times 55.85 \mathrm{~g} / \mathrm{mol}$

$=0.205 \mathrm{~g}$

Find $\% \mathrm{Fe}(\mathrm{II})$ in unknown sample

$\begin{aligned}\% \mathrm{Fe} &=(\text { mass of } \mathrm{Fe} / \text { weight of unknown sample }) \times 100 \\&=(0.205 \mathrm{~g} / 2.4234 \mathrm{~g}) \times 100 \\&=8.46 \%\end{aligned}$

Molarity is the concentration of a solution expressed as the number of moles of solute dissolved in each liter of solution.

concentration is the amount of a substance per defined space. Concentration usually is expressed in terms of mass per unit volume.

To know more about MOLARITY visit : brainly.com/question/8732513

#SPJ4

7 0

2 years ago

Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 83 vol% carbide

velikii [3]

Answer:

k_max = 31.82 w/mk

k_min = 17.70 w/mk

Explanation:

a) the maximum thermal conductivity is given as

$k_{max} = k_m*v_m + k_p*v_p$

where k_m is thermal conductvitiy of metal

k_p is thermal conductvitiy of carbide

v_m = proportion of metal in the cement = 0.17

v_p = proportion of carbide in the cement = 0.83

= 66*0.17 + 28*0.83

k_max = 31.82 w/mk

b) the minimum thermal conductivity is given as

$k_{min} = \frac{k_{carbide}* k_{metal}}{k_{metal} v_{carbide} +k_{metal} v_{carbide}}$

= \frac{28+66}{28*0.17 +66*0.83}

k_min = 17.70 w/mk

7 0

3 years ago

If you want to stop the current flow through Device 3 in the circuit shown above, which one of the following single switches sho

Arturiano [62]

Thats a hard one, From The looks of it, it looks like B, Because it would leave a gap in the flow for a small amount of time

7 0

4 years ago

Read 2 more answers

Gauss's Law states that the net electric flux, , through any closed surface is proportional to the charge enclosed: . The analog

Natali [406]

The analogous formula for magnetic fields is the Ampere's law.

To find the answer, we need to know about the Ampere's law of magnetism.

<h3>What's Ampere's law of magnetism?</h3>

Ampere's law states that the close line integral of magnetic field around a current carrying loop is directly proportional to the current enclosed within it.

<h3>What's is the mathematical expression of Ampere's law?</h3>

Mathematically, Ampere's law is

B•dl= μ₀I

Thus, we can conclude that the analogous formula for gauss law is the Ampere's law in magnetism.

Learn more about the Ampere's law here:

brainly.com/question/17070619

#SPJ4

5 0

3 years ago

uppose that the terminal speed of a particular sky diver is 150 km/h in the spread-eagle position and 320 km/h in the nosedive p

AysviL [449]

Answer:

4.55

Explanation:

The terminal speed of a diver is given by:

$v_t=\sqrt{\frac{2mg}{C\rho A} } \\\\Where\ m=mass\ of \ driver,d=acceleration\ due\ to\ gravity,C=drag\ \\coefficient,A=cross\ sectional\ Area.\\\\Therefore:\\\\A=\frac{2mg}{C \rho v_t^2} \\\\For\ area\ with\ terminal\ speed\ in\ spread\ angle\ position(v_s):\\\\A_s=\frac{2mg}{C \rho v_s^2} \\\\For\ area\ with\ terminal\ speed\ in\ nose\ dive\ position(v_n):\\\\A_n=\frac{2mg}{C \rho v_n^2}\\\\Therefore\ since\ g,m,C,\rho\ are\ constant:\\\\$

$\frac{A_s}{A_n}= \frac{\frac{2mg}{C \rho v_s^2}}{\frac{2mg}{C \rho v_n^2}}\\\\\frac{A_s}{A_n}= \frac{v_n}{v_s} \\\\v_n=320\ km/h,v_s=150\ km/h\\\\\frac{A_s}{A_n}=\frac{320^2}{150^2} =4.55$

4 0

3 years ago