Which of the following is not true of alloys?

How does an Arrhenius acid change when it<br> dissolves in water?

salantis [7]

Answer:

Explanation:

when dissolved in water, dissociates into ions, including hydrogen ions. According to Arrhenius, an acid can be defined as a material that increases the concentration of hydrogen ions in water.

5 0

3 years ago

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157g of the compound produced 0.213g of CO2 and

vladimir2022 [97]

Answer:

$C_{7} H_{5}N_{3}O_{6}$

Explanation:

First reaction gives you the number of moles or the mass from Carbon and hydrogen

for carbon:

$0,213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} =0.005molC$

$0.213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} .\frac{12gC}{1molC} = 0.058gC\\$

Analogously for hydrogen:

0.0310g $H_{2}O$ have 0.0034gH or 0.0034mol of H

In the second reaction you can obtain the amount of nitrogen as a percentage and find the mass of N in the first sample.

$0.023gNH_{3} .\frac{1molNH_{3}}{17gNH_{3}} .\frac{1molN}{1molNH_{3}} .\frac{14gN}{1molN} \frac{100}{0.103gsample} =18.4%N$

now

$\frac{18.4gN}{100gsample} .0.157gsample=0.0289gN in the first reaction$

this is equivalet to 0.002mol of N

with this information you can find the mass of oxygen by matter conservation.

$gO=total mass-(gN+gC+gH)=0.157-(0.0289+0.058+0.0034)=0.0666gO$

this is equivalent to 0.004molO

finally you divide all moles obtained between the smaller number of mole (this is mol of H)

$C\frac{0.0048}{0.0034} H\frac{0.0034}{0.0034} N\frac{0.002}{0.0034} O\frac{0.004}{0.0034} =C_{1.4} HN_{0.6} O_{1.2}$

and you can multiply by 5 to obtain: $C_{7} H_{5}N_{3}O_{6}$

4 0

3 years ago

How can you test a mineral for hardness?<br><br> I NEED~ THIS NOW PLEASE

creativ13 [48]

Courtney’s in minerals can vary due to impurities but is usually diagnostic we determine the relative hardness of minerals using a scale diversity bye mineralogists Freidrich Mohs The scale assigned hardness Tuten common index minerals and is based up on the ability of one mineral to scratch another

8 0

3 years ago

Enter a chemical equation for NaOH(aq) showing how it is an acid or a base according to the Arrhenius definition. Consider that

yuradex [85]

Answer:

NaOH(aq) is a Base.

Explanation:

Those substances which give or release $OH^{-}$ ions in aqueous solution are called as the Arrhenius Bases.

In the aqueous solution, NaOH dissociates as follows -

$NaOH^{}$ ↔ $Na^{+} + OH^{-}$

If it reacts with a strong acid HCl, the chemical equation for this reaction will be as follows -

$HCl + NaOH = NaCl + H_{2} O$

6 0

2 years ago

Consider the reaction: P4 + 6Cl2 = 4PCl3.

likoan [24]

Answer:

The answer to your question is below

Explanation:

Consider the reaction: P4 + 6Cl2 = 4PCl3.

a. How many grams of Cl2 are needed to react with 20.00 g of P4? ___68.7 g___________

P4 + 6Cl2 = 4PCl3

4(31) ---------- 12(35.5)

20 ---------- x

x = 20(12x35.5) / 4(31)

x = 8520 / 124

x = 68.7 g

b. You have 15.00 g. of P4 and 22.00 g. of Cl2, identify the limiting reactant and calculate the grams of PCl3 that can be produced as well as the grams of excess reactant remaining. LR____________ grams PCl3 _________ grams excess reactant ___________

P4 + 6Cl2 = 4PCl3

124g 426 g 4(31 + 3(35.5)) = 550g

15g 22g

I will use P4 to find the limiting reactant

x = (15 x 426) / 124 = 51.5 The limiting reactant is Chlorine

because we need 51.5 g and we only have 22g

Excess reactant

x = (22 x 124) / 426 = 6.4 g of P4

Excess P4 = 15 g - 6.4 = 8.6 g of P4 in excess

Grams of PCl3 produced

426 g of Cl2 ---------------- 550 g of PCl3

22g of Cl2 ------------- - x

x = (22 x 550) / 426 = 28.4 g of PCl3

c. If the actual amount of PCl3 recovered is 16.25 g., what is the percent yield? ______________

% yield = (16.25 - 28.4) / 28.4 x 100

% yield = 42.8

d. Given 28.00 g. of P4 and 106.30 g. of Cl2, identify the limiting reactant and calculate how many grams of the excess reactant will remain after the reaction. LR ______________ grams excess reactant

Limiting reactant

124g of P4 ------------- 426 g 6Cl2

28g --------------- x

x = (28 x 426) / 124

x = 96.2 g of Cl2 and we have 106.3 so Chlorine is the excess reactant and P4 is the limiting reactant.

Excess reactant = 106.3 - 96.2 = 10.1 g of Cl2 in excess

4 0

2 years ago