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solniwko [45]
3 years ago
8

A sample of metal had a mass of 17.29g and a volume of 5.68mL what is the density of this metal

Chemistry
1 answer:
Gennadij [26K]3 years ago
8 0
Density = mass/volume

<span>17.29g / 5.68mL = 3.044g/mL</span>
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We dissolve 93 grams of sodium sulfate (Na2SO4) in 20 grams of water. What is the percent by mass of sodium sulfate in this solu
dlinn [17]
It should be for the total solution of 93 plus 20 grams which is 113 grams so 93 divided by 113 grams comes to 82.3% sodium sulfate and this can be checked by multiplying 113 grams by 0.823 which results in 93 grams of sodium sulphate. 
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Consider the chemical equation below.
Aleks [24]
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6 0
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05. When a gold pebble is placed in a graduated cylinder that contains 12.0 mL of water, the water level rises
OLga [1]

Answer:

142.82 g

Explanation:

The following data were obtained from the question:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Density of gol= 19.3 g/cm³

Mass of gold =.?

Next, we shall determine the volume of the gold. This can be obtained as follow:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Volume of gold =.?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 19.4 – 12

Volume of gold = 7.4 mL

Finally, we shall determine the mass of the gold as follow:

Note: 1 mL is equivalent to 1 cm³

Volume of gold = 7.4 mL

Density of gol= 19.3 g/cm³ = 19.3 g/mL

Mass of gold =?

Density = mass /volume

19.3 = mass of gold /7.4

Cross multiply

Mass of gold = 19.3 × 7.4

Mass of gold = 142.82 g

Therefore, the mass of the gold pebble is 142.82 g

6 0
3 years ago
The extend of sewage treatment required mostly depends on the (A) Control of suspended solid (B) BOD of the sewage (C) Both a an
romanna [79]

The extent of sewage treatment required mostly depends on the control of suspended solids and BOD of the sewage. Hence, option C is correct.

<h3>What is Inorganic matter?</h3>

Inorganic matter is substances of mineral origin that are not characterized by primarily carbon-based structures.

Sewage treatment (or domestic wastewater treatment, municipal wastewater treatment) is a type of wastewater treatment which aims to remove contaminants from sewage to produce an effluent that is suitable for discharge to the surrounding environment or an intended reuse application, thereby preventing water pollution from raw sewage discharges.

The main goal of wastewater treatment facilities is to protect humans and the ecosystem from harmful and toxic elements found in wastewater.

Hence, option C is correct.

Learn more about the Inorganic matter here:

brainly.com/question/17163874

#SPJ1

8 0
1 year ago
Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemogl
Alex73 [517]

Without wasting much of our time, Here is the correct question.

Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen hemoglobin(aq) + O2(aq) -------> hemoglobin O2(aq) is first order in hemoglobin and first order in dissolved oxygen, with a rate constant of 4 × 10⁷ L mol⁻¹ s⁻¹. Calculate the initial rate at which oxygen will be bound to hemoglobin if the concentration of hemoglobin is 2 × 10⁻⁹ M and that of oxygen is 5 × 10⁻⁵M.

Answer:

4 × 10⁻⁶ M s⁻¹

Explanation:

The equation for the reaction between Hemoglobin molecules in blood that binds with oxygen molecule can be represent by:

hemoglobin_{(aq)  +  O_{2(aq)   ---------> hemoglobin.O_{2(aq)

Now, we are also being told to calculate only!, the  initial rate at which oxygen will be bound to hemoglobin.

So, If it is first order in hemoglobin and also first order in Oxygen molecule at the initial rate of the the reaction, therefore, the rate  for the reaction can be expressed as :

rate = k [hemoglobin_{(aq)}][O_{2(aq)}]

Let's not forget that we are so given some parameters;

where

k (rate constant) = 4 × 10⁷ L mol⁻¹ s⁻¹

[ hemoglobin_{(aq) ] = 2 × 10⁻⁹ M

[  O_{2(aq)  ]  =  5 × 10⁻⁵ M

Substituting our data given into the above rate formula, we have:

rate = (4 × 10⁷ L mol⁻¹ s⁻¹) × (2 × 10⁻⁹ M) × (5 × 10⁻⁵ M)

rate = 4 × 10⁻⁶ M s⁻¹     ( given that 1 M = 1 mol L⁻¹ )

∴ the initial rate at which oxygen will be bound to hemoglobin = 4 × 10⁻⁶ M s⁻¹

7 0
3 years ago
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