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2 years ago

What is the element found in Group 17 Period 4?

Chemistry

1 answer:

faust18 [17]2 years ago

3 0

Answer:

Bromine is found in Group 17, Period 4 on the periodic table.

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What volume of 0.305 m agno3 is required to react exactly with 155.0 ml of 0.274 m na2so4 solution? hint: you will want to write

LuckyWell [14K]

The balanced chemical equation for reaction of $AgNO_{3}$ and $Na_{2}SO_{4}$ is as follows:

$2 AgNO_{3}+Na_{2}SO_{4}\rightarrow 2NaNO_{3}+Ag_{2}SO_{4}$

From the balanced chemical equation, 2 mol of $AgNO_{3}$ reacts with 1 mol of $NaNO_{3}$ .

First calculating number of moles of $NaNO_{3}$ as follows:

$M=\frac{n}{V}$

On rearranging,

$n=M\times V$

Here, M is molarity and V is volume. The molarity of $NaNO_{3}$ is given 0.274 M or mol/L and volume 155 mL, putting the values,

$n=0.274 mol/L\times 155\times 10^{-3}mL=0.04247 mol$

Since, 1 mol of $NaNO_{3}$ reacts with 2 mol of $AgNO_{3}$ thus, number of moles of $AgNO_{3}$ will be $2\times 0.04247 mol=0.08494 mol$ .

Now, molarity of $AgNO_{3}$ is given 0.305 M or mol/L thus, volume can be calculated as follows:

$V=\frac{n}{M}=\frac{0.08494 mol}{0.305 mol/L}=0.2785 L=278.5 mL$

Therefore, volume of $AgNO_{3}$ is 278.5 mL.

4 0

3 years ago

If 35 ml of 6.0 m h2so4 was spilled, calculate the minimum mass of nahco3 that must be added to the spill to neutralize the acid

belka [17]

The balanced equation between the $H_2SO_4$ and $NaHCO_3$ is:

$H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O$

Formula of molarity is:

$Molarity = \frac{Moles of solute}{Volume of solution in Liters}$

$Molarity = 6.0 M, Volume = 35 mL = 0.035 L$

Substituting the values,

$6 = \frac{Moles of solute}{0.035}$

$Moles of solute = 0.035 L\times 6 mol/L = 0.21 mole$

So, number of moles of $H_2SO_4$ is $0.21 mole$ .

From the balanced equation it is clear that for 1 mole of $H_2SO_4$ , 2 moles of $NaHCO_3$ are required.

Hence, 0.21 mole of $H_2SO_4$ = $2\times 0.21 mole = 0.42 mole of NaHCO_3$

Molar mass of $NaHCO_3$ = $84.007 g/mol$

So, the mass of $NaHCO_3$ = $84.007 g/mol \times 0.42 mol = 35.283 g$

6 0

2 years ago

A balloon inflated in a room at 27 degrees celsius has a volume of 8.00 L.the balloon is then heated to a temperature of 78 degr

dangina [55]

This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 27 °C = 300 K
Initial Pressure = P₁ = constant
Initial Volume = V₁ = 8 L
Final Temperature = T₂ = 78 °C = 351 K
Final Pressure = P₂ = constant
Final Volume = V₂ = ?
As,
Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
V₁ / T₁ = V₂ / T₂
Solving for V₂,
V₂ = (V₁ × T₂) ÷ T₁
Putting Values,
V₂ = (8 L × 351 K) ÷ 300 K

V₂ = 9.38 L

5 0

2 years ago

The surface temperatures on land vary more over the course of a year than the surface temperatures on the ocean because land has

bogdanovich [222]

Answer:

Lower heat capacity

Explanation:

The heat or thermal capacity is a physical property defined as the amount of heat a material need in order to elevate a unit in its temperature, this means that water increases its temperature more easily than land.

I hope you find this information useful and interesting! Good luck!

7 0

2 years ago

Not sure what to do can you help

NeTakaya

sorry i don't know the answer i'm really sorry

7 0

2 years ago

What is the element found in Group 17 Period 4?​

What is the element found in Group 17 Period 4?