The only reactants in a reaction were Zn and HCl. What does the law of

Why might an idea or hypothesis be discarded? A. If another scientist doesn't like it. B. If evidence also supports another hypo

vladimir1956 [14]

Answer:

C. If it is tested and the evidence does not support it.

Explanation:

A hypothesis is more less a scientific guess. Before such a guess or prediction is made, empirical observations and deductions are first made. It is from the result of the observations that a hypothesis statement is made.

For a hypothesis to become widely adopted and accepted, it must be testable within the limits of the experiment as described by the proposer. When subjected to test and it agrees, the status of a hypothesis can be upgraded.

If the hypothesis is tested and evidence contrasts the result being sort for, a hypothesis will be discarded.

7 0

3 years ago

Elemental mercury combines with oxygen to make mercury oxide. write a balanced equation

Ilia_Sergeevich [38]

Hg + O2 --> HgO + O

Note that Mercury has a variable valency.

6 0

3 years ago

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

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

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0

3 years ago

List the number of each type of atom on the left side of the equation 2C10H22(l)+31O2(g)→20CO2(g)+22H2O(g)

Leokris [45]

Answer:

20, 44, 62

Explanation:

To find the number of atoms of each element, we multiply coefficient and subscript

For example $5 Ca_1 Cl_2$ contains

5 × 1 = 5 ,Ca atoms and

5 × 2 = 10, Cl atoms

If there is a bracket in the chemical formula

For example $3Ca_3 (P_1 O_4 )_2$

we multiply coefficient × subscript × number outside the bracket to find the number of atoms

(Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

So

3 × 3 = 9 , Ca atoms

3 × 1 × 2 = 6, P atoms

3 × 4 × 2 = 24, O atoms are present.

So let us find the number of atoms of each element on the left side of the equation

$2C_{10} H_{22} (l)+31O_2 (g)\Rightarrow 20CO_2 (g)+22H_2 O(g)$

Number of C atoms = 2 × 10 = 20

Number of H atoms = 2 × 22 = 44

Number of O atoms = 31 × 2 = 62

20, 44, 62 are the Answers.

3 0

3 years ago

How might the biodiversity of a mowed lawn compare to that of huge weedy field?

Dennis_Churaev [7]

Answer: The mowed lawn is the one from where the grasses are removed by using the machines or tools.

Explanation:

The mowed lawn is expected to have low number of species as the grasses may be few or scanty thus can support the population of few species like insects, mice, birds, and small number of grazing animals. On the other hand the weedy field can be hub of insects, reptiles like snakes, small mammals, and large mammals. Large weed field can provide food, and habitat to the large number of species. This will support the increase in biodiversity as compared to the mowed lawn.

5 0

3 years ago