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diamong [38]
3 years ago
13

1. Water is easily contaminated by both biological and chemical contaminants. Define and give 3 examples of each

Chemistry
1 answer:
tamaranim1 [39]3 years ago
7 0
Are there options like A,B,C,D?

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A base
hichkok12 [17]
B, turns red litmus paper to blue
5 0
3 years ago
To study earths interior, geologist often rely on indirect methods, such as evidence from fossils.
igomit [66]

Answer: false

Explanation:

It is false that to study Earth's interior, geologists often rely on indirect methods, such as evidence from fossils. They rely on seismic wave.

6 0
3 years ago
Read 2 more answers
What 2 chemicals react together during photosynthesis
Nitella [24]

Answer:

The photosynthesis chemical equation states that the reactants (carbon dioxide, water and sunlight), yield two products, glucose and oxygen gas. The single chemical equation represents the overall process of photosynthesis.

Explanation:

Definition: the process by which green plants and some other organisms use sunlight to synthesize foods from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a byproduct.

Hope this helps!

Can I get brainliest?

5 0
3 years ago
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
What volume of a 0.155 M potassium hydroxide solution is required to neutralize 25.7 mL of a 0.388 M hydrobromic acid solution
vekshin1

Answer: Therefore, the volume of a 0.155 M potassium hydroxide solution  is 56.0 ml

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

M_1 = molarity of HBr solution = 0.338 M

V_1 = volume of HBr solution = 25.7 ml

M_2 = molarity of KOH solution = 0.155 M

V_2 = volume of KOH solution = ?

n_1 = valency of HBr = 1

n_2 = valency of KOH = 1

1\times 0.338\times 25.7=1\times 0.155\times V_2

V_2=56.0ml

Therefore, the volume of a 0.155 M potassium hydroxide solution  is 56.0 ml

8 0
3 years ago
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