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2 years ago

Which item is made from an alloy? A. steel tray B. glass plate C. credit card D. copper wire

Chemistry

2 answers:

Naya [18.7K]2 years ago

6 0

Not sure if this is correct but i would choose the steel tray.

Anna007 [38]2 years ago

4 0

It's A).steel tray because steel is an iron-carbon alloy.

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Is aloe vera a living or non-living thing?

Afina-wow [57]

Answer:

It is a living plant

Explanation:

3 0

2 years ago

What is the stoichiometric coefficient for oxygen when the following equation is balanced using the lowest whole-number coeffici

rodikova [14]

Answer:

Option (B) 7

Explanation:

C3H6O2(l) + O2(g) → CO2(g) + H2O(l)

To know the coefficient of O2 in the above equation, let us balance the equation.

The above equation can be balance as follow:

C3H6O2(l) + O2(g) → CO2(g) + H2O(l)

There are 3 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 3 in front of CO2 as shown below:

C3H6O2(l) + O2(g) → 3CO2(g) + H2O(l)

There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

C3H6O2(l) + O2(g) → 3CO2(g) + 3H2O(l)

There are a total of 4 atoms of O on the left side and a total of 9 atoms on the right side. It can be balance by putting 7/2 in front of O2 as show below:

C3H6O2(l) + 7/2O2(g) → 3CO2(g) + 3H2O(l)

Multiply through by 2

2C3H6O2(l) + 7O2(g) → 6CO2(g) + 6H2O(l)

Now, the equation is balanced.

From the balanced equation above, the coefficient of O2 is 7.

7 0

2 years ago

Convert 675 nanograms to milligrams.

STALIN [3.7K]

675 nanograms is equivalent to 6.75 × 10-⁴ milligrams.

<h3>How to convert nanograms to milligrams?</h3>

Nanograms is a unit of mass equal to 0.000000001 grams and has a symbol of ng.

On the other hand, milligrams is another unit of mass equal to 0.001grams and has a symbol of mg.

Nanograms and milligrams are both units of mass and can be inter-convertible as follows:

1 nanogram = 1 × 10-⁶ milligram

According to this question, 675 nanograms is equivalent to 675 × 10-⁶ milligrams.

Therefore, 675 nanograms is equivalent to 6.75 × 10-⁴ milligrams.

Learn more about milligrams at: brainly.com/question/20320382

#SPJ1

7 0

1 year ago

The decomposition of HBr(g) into elemental species is found to have a rate constant of 4.2 ×10−3atm s−1. If 2.00 atm of HBr are

Dennis_Churaev [7]

Answer:

7,94 minutes

Explanation:

If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the reaction rate does not depend on the concentration of the reactants.

For the zero-order reactions, concentration-time equation can be written as follows:

[A] = - Kt + [Ao]

where:

[A]: concentration of the reactant A at the t time,
[A]o: initial concentration of the reactant A,
K: rate constant,
t: elapsed time of the reaction

To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.

Data:

K = 4.2 ×10−3atm/s,

[A]o=[HBr]o= 2 atm,

[A]=[HBr]=0 atm (all HBr(g) is gone)

We clear the incognita :

[A] = - Kt + [Ao]............. Kt = [Ao] - [A]

t = ([Ao] - [A])/K

We replace the numerical values:

t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes

So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).

6 0

2 years ago

The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A

s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $m^3 /d$

$= 80 \times 10^5 \ I/d$

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L$

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d$

If 60 mg of alum $[Al_2(SO_4)_3.14 H_2O]$ required for one litre of the water treatment.

So Alum required for $80 \times 10^5 \ I/d$

$= 80 \times 15^5 \ I/d \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d$

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $= 60 \times 0.234 \text{ of alum ppt. per litre}$

$= 14.04$ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

= 112.32 kg/d ppt of alum

Daily total solid load is $= 144 \ kg/d + 112.32 \ kg/d$

= 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234$

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$

After addition, the aluminium hydroxide pH of water will increase due to increase in $OH^-$ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm$

And the pH is reduced into the range of 5.9 to 6.4

6 0

2 years ago