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2 years ago

Which of the following represents the relative charge of proton

Chemistry

1 answer:

Snowcat [4.5K]2 years ago

8 0

You’ve not put anything below so I’m not sure what the options are but the relative charge of a proton is +1

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Please anser this asap

Fittoniya [83]

Answer:

14 is the 1 st one and the 15 is 3

8 0

2 years ago

Read 2 more answers

A military jet cruising at an altitude of 12.0kmand speed of 1900./kmh burns fuel at the rate of 74.4/Lmin. How would you calcul

anygoal [31]

Answer: amount = 2466.95L

Explanation:

given that the speed is = 1900./kmh i.e. 1hr/900km

distance = 1050km

the fuel burns at a rate of 74.4 L/min

therefore the amount of fuel that the jet consumes on a 1050.km becomes;

total fuel used = time × fuel burning rate

where time = distance / speed

∴ total fuel used (consumed) = time × fuel burning rate

total fuel consumed = (1050km × 1hr/1900km) × (60min/ 1hr × 74.4L/1min)

total fuel consumed = 2466.95L

6 0

3 years ago

Periods On the periodic table represent elements

Natalka [10]

Answer:When you look at the periodic table, each row is called a period (Get it? Like PERIODic table.). All of the elements in a period have the same number of atomic orbitals. For example, every element in the top row (the first period) has one orbital for its electrons.

Explanation:

7 0

2 years ago

This element is used to improve the conductivity of germanium.

Ber [7]

Gallium is used to improve the conductivity of germanium

3 0

2 years ago

A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$