What is the chemical formula for hydrogen peroxide?

Which sediment type requires the least amount of energy to be eroded

castortr0y [4]

Answer:

SEDIMENTARY ROCK

Explanation:

8 0

4 years ago

In part a, we saw that the theoretical yield of aluminum oxide is 1.40 mol . calculate the percent yield if the actual yield of

Goryan [66]

<span>1.4 moles of aluminium metal is exposed to 1.35 mol of oxygen Theoretical yield=0.007 mol calculate % yield % yield= actual yield/theoretical yield*100 % yield= 0.938/0.700*100 % yield= 13.4% yield</span>

7 0

3 years ago

Reactivity and flammability are are examples of properties.

nika2105 [10]

Answer:

they are eamples of chemical properties

Explanation:

flammibility is the chemical makeup of the element that makes it flamable and reactivitie is the chemical makeup of an element that is able to reacte to another element

3 0

3 years ago

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are

vaieri [72.5K]

Answer: Resulting solution will not be neutral because the moles of $OH^-$ ions is greater. The remaining concentration of $[OH^-]$ ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

$[HNO_3]$ = 0.200 M

$[Ca(OH)_2]$ =0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of $H^+$ ion from both the acids

moles of $H^+$ in HCl

$HCl\rightarrow {H^+}+Cl^-$

if 1 L of $HCl$ solution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 $\times$ 0.1 moles= 0.005 moles (1L=1000mL)

moles of $H^+$ in HCl = 0.005 moles

Similarliy

moles of $H^+$ in $HNO_3$

$HNO_3\rightarrow H^++NO_3^-}$

If 1L of $HNO_3$ solution= 0.200 moles

Then 0.1L of $HNO_3$ solution= 0.1 $\times$ 0.200 moles= 0.02 moles

moles of $H^+$ in $HNO_3$ =0.02 moles

so, Total moles of $H^+$ ions = 0.005+0.02= 0.025 moles .....(1)

Step 2: Calculating the total moles of $[OH^-]$ ion from both the bases

Moles of $OH^-\text{ in }Ca(OH)_2$

$Ca(OH)_2\rightarrow Ca^2{+}+2OH^-$

1 L of $Ca(OH)_2$ = 0.0100 moles

Then in 0.5 L $Ca(OH)_2$ solution = 0.5 $\times$ 0.0100 moles = 0.005 moles

$Ca(OH)_2$ produces two moles of $OH^-$ ions

moles of $OH^-$ = 0.005 $\times$ 2= 0.01 moles

Moles of $OH^-$ in $RbOH$

$RbOH\rightarrow Rb^++OH^-$

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2 $\times$ 0.100 moles = 0.02 moles

Moles of $OH^-$ = 0.02 moles

so,Total moles of $OH^-$ ions = 0.01 + 0.02=0.030 moles ....(2)

Step 3: Comparing the moles of both $H^+\text{ and }OH^-$ ions

One mole of $H^+$ ions will combine with one mole of $OH^-$ ions, so

Total moles of $H^+$ ions = 0.005+0.02= 0.025 moles....(1)

Total moles of $OH^-$ ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of $H^+$ ions = total moles of $OH^-$ ions

0.025 moles $H^+$ will neutralize the 0.025 moles of $OH^-$

Moles of $OH^-$ ions is in excess (from 1 and 2)

The remaining moles of $OH^-$ will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of $OH^-$ ions = $\frac{\text{Moles remaining}}{\text{Total volume}}$

$[OH^-]=\frac{0.005}{0.85}=0.0058M$

6 0

4 years ago

Fill in the blank with the correct answer to complete the sentence.

guajiro [1.7K]

Explanation:

momentum increases with an increase in mass and velocity

3 0

3 years ago