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2 years ago

ILL GIVE U 100 POINTS!!!!

Chemistry

2 answers:

Alexxx [7]2 years ago

6 0

✿I'll take that 50 pts ✿

✨Refer these attachments, for your answer ✨

hope it helps ❤~

✨luv, Snowflake✨

MrMuchimi2 years ago

4 0

Answer:

Explanation:

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What is a half-life in Chemistry?

makvit [3.9K]

Answer:

half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive ...

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2 years ago

A pharmacist wishes to strengthen a mixture from 10%alcohol to 30% alcohol. How much pure alcohol should be added to 7 liters of

Tomtit [17]

Answer:

2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.

Explanation:

Suppose x is the number of litres added to the 10% mixture than the quantity of new mixture is given as below

$n_{old}=7$ litres
$n_{new}=7+x$ litres

Also the quantity of alcohol is given as

$q_{old}=10 \% \, of \, 7 \, litres =0.7$

$q_{added}=x$

$q_{new}= 30 \% \,of \,new\, quantity = 0.3(7+x)$

Now the equation is as

$q_{old}+q_{added}=q_{new}\\0.7+x=0.3(7+x)\\0.7+x=2.1+0.3x\\x-0.3x=2.1-0.7\\0.7x=1.4\\x=2 \, litres$

So 2 litres of pure alcohol will be added to make the overall concentration of 9 litres of mixture as 30%.

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3 years ago

When describing velocity what should be included?

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3 years ago

Read 2 more answers

how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate

Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

$3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3$

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

$0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})$

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

$63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})$

= $95.7gNaClO_3$

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate and then the grams are converted to kg.

$95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})$

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

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3 years ago

Element

Katyanochek1 [597]

Answer:

the two elements are in the same period, with element R the first element in the period and element Q the last element

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