What period are the following elements in He,Ge,Rb,l

Suppose 17. g of hydrochloric cid is mixed with 6.99 g of sodium hydroxide calculate the minimum mass of hydochloric acid taht c

Eva8 [605]

Answer: 10.62g

Explanation:

First let us generate a balanced equation for the reaction.

HCl + NaOH —> NaCl + H2O

Molar Mass of HCl= 1 + 35.5 = 36.5g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

From the question,

Mass of HCl = 17g

Mass of NaOH = 6.99g

Converting these Masses to mole, we obtain:

n = Mass / Molar Mass

n of HCl = 17/36.5 = 0.4658mol

n of NaOH = 6.99/40 = 0.1748mol

From the question,

1 mole of NaOH requires 1mole of HCl.

Therefore, 0.1748mol of NaOH will also require 0.1748mol of HCl.

But we were told that 17g( i.e 0.4658mol) of HCl were mixed.

Therefore, the unreacted amount of HCl = 0.4658 — 0.1748 = 0.291mol

Converting this to mass, we have:

Mass of HCl = n x molar Mass

Mass of HCl = 0.291 x 36.5

Mass of HCl = 10.62g

Therefore the left over Mass of HCl is 10.62g

8 0

3 years ago

If 15.6 grams of copper (ii) chloride react with 20.2 grams of sodium nitrate how many grams of sodium chloride can be formed? W

olasank [31]

Answer:

- 13.56 g of sodium chloride are theoretically yielded.

- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.

- 0.50 g of sodium nitrate remain when the reaction stops.

- 92.9 % is the percent yield.

Explanation:

Hello!

In this case, according to the question, it is possible to set up the following chemical reaction:

$CuCl_2+2NaNO_3\rightarrow 2NaCl+Cu(NO_3)_2$

Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:

$m_{NaCl}^{by\ CuCl_2}=15.6gCuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} *\frac{2molNaCl}{1molCuCl_2} *\frac{58.44gNaCl}{1molNaCl} =13.56gNaCl\\\\m_{NaCl}^{by\ NaNO_3}=20.2gNaNO_3*\frac{1molNaNO_3}{84.99gNaNO_3} *\frac{2molNaCl}{2molNaNO_3} *\frac{58.44gNaCl}{1molNaCl} =13.89gNaCl$

Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:

$m_{NaNO_3}^{by\ NaCl}=13.56gNaCl*\frac{1molNaCl}{58.44gNaCl}*\frac{2molNaNO_3}{2molNaCl} *\frac{84.99gNaNO_3}{1molNaNO_3}=19.72gNaNO_3$

Therefore, the leftover of sodium nitrate is:

$m_{NaNO_3}^{leftover}=20.2g-19.7g=0.5gNaNO_3$

Finally, the percent yield is computed via:

$Y=\frac{12.6g}{13.56g} *100\%\\\\Y=92.9\%$

Best regards!

6 0

3 years ago

Mr. Bullis wants to know if he gives teachers candy, it will make them smile more. He gives half the teachers candy and the othe

mote1985 [20]

Answer:

the experimental group would be the teachers who recieved the candy

control group would be the teachers who dont get any candy

<em>Explanation:</em>

experimental group is the group that receives the test variable being tested in this case the variable being the candy

and the control group is the group who doesn't receive the variable

purpose of having a control is to rule out other factors which may influence the results of an experiment.

5 0

3 years ago

when an action potential abides by the all-or-nothing principle, once it reaches its threshold it moves all the way down the axo

Karolina [17]

Answer:

Yep and it's threshold is -55 mV.

3 0

3 years ago

If you had 10.00 mL of acetic acid that you titrated with .100 M NaOH, and you used 45.15 mL of that base to get a light pink co

3241004551 [841]

Answer:

0.4515 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = ? M

V₁ = 10.0 mL

M₂ = 0.100 M

V₂ = 45.15 mL

Using the above formula , the molarity of acid , can be calculated as ,

M₁V₁ = M₂V₂

M₁ * 10.0mL = 0.100 M * 45.15 mL

M₁ = 0.4515 M

Hence, the concentration of the acetic acid = 0.4515 M

4 0

3 years ago