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Alexxx [7]
3 years ago
5

When patterning a shotgun, what is a sufficient percentage of pellets within a 30-inch circle? at least 25% of the load, concent

rated in the center at least 35% of the load, with even distribution at least 45% of the load, concentrated in the center at least 55% of the load, with even distribution?
Chemistry
1 answer:
Murrr4er [49]3 years ago
3 0
<span>According to state hunter's education standards, your shotgun pattern within a 30 inch circle should be concentrated in the center at least 55% of the load, with even distribution.</span>
You might be interested in
Given that the formula of butane C4H10 the accepted value for the molar mass should be
deff fn [24]

Answer:

carbon mass = 12.01g/mol

hydrogen mass = 1.01g/mol

4 carbon atoms and 10 hydrogen so

12.01 x 4 + 1.01 x 10

48.04g/mol + 10.10g/mol

= 58.14g/mol

4 0
3 years ago
If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
Write the chemical formula for sodium hypobromite.
irga5000 [103]

Answer:

             NaOBr  (or)  Na⁺ ⁻OBr

Explanation:

The Oxo-Acids of Bromine are as follow,

                           Hypobromous Acid  =  HOBr

                           Bromous Acid  =  HOBrO

                           Bromic Acid  =  HBrO₃

                           Perbromic Acid  =  HBrO₄

When these acids are converted to their conjugate bases their names are as follow,

                           Hypobromite  =  ⁻OBr

                           Bromite  =  ⁻OBrO

                           Bromate  =  ⁻OBrO₂

                           Perbromate  =  ⁻OBrO₃

According to rules, the positive part of ionic compound is named first and the negative part is named second. So, Sodium Hypobromite has a chemical formula of Na⁺ ⁻OBr or NaOBr.

6 0
3 years ago
Collected data consists of:
san4es73 [151]

The number of moles of the magnesium (mg) is 0.00067 mol.

The number of moles of hydrogen gas is 0.0008 mol.

The volume of 1 more hydrogen gas (mL) at STP is 22.4 L.

<h3>Number of moles of the magnesium (mg)</h3>

The number of moles of the magnesium (mg) is calculated as follows;

number of moles = reacting mass / molar mass

molar mass of magnesium (mg) = 24 g/mol

number of moles = 0.016 g / 24 g/mol = 0.00067 mol.

<h3>Number of moles of hydrogen gas</h3>

PV = nRT

n = PV/RT

Apply Boyle's law to determine the change in volume.

P1V1 = P2V2

V2 = (P1V1)/P2

V2 = (101.39 x 146)/(116.54)

V2 = 127.02 mL

Now determine the number of moles using the following value of ideal constant.

R = 8.314 LkPa/mol.K

n = (15.15 kPa x 0.127 L)/(8.314 x 290.95)

n = 0.0008

<h3>Volume of 1 mole of hydrogen gas at STP</h3>

V = nRT/P

V = (1 x 8.314 x 273) / (101.325)

V = 22.4 L

Learn more about number of moles here: brainly.com/question/13314627

#SPJ1

7 0
2 years ago
PLEASE HELP ITS DUE TODAY
dimulka [17.4K]

Answer:

2nd one and last one

Explanation:

im just that smart

8 0
3 years ago
Read 2 more answers
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