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3 years ago

What is the state of water at room temperature? Please someone help thank you

Chemistry

2 answers:

velikii [3]3 years ago

7 0

it is in a liquid state

in room temperature

Arlecino [84]3 years ago

4 0

Answer:

liquid state

Explanation:

water is liquid at room temperature because of the many weak hydrogen bonds that holds its molecules together in small fractions of second

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Which of the following is the correct sequence for the discovery of subatomic particles? a Electrons> Protons > Neutrons b

laila [671]

Answer:

Explanation;

Maybe is C

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3 years ago

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A metal oxide with the formula mo contains 15.44% oxygen. in the box below, type the symbol for the element represented by m.

lana66690 [7]

<u>Answer:</u> The element represented by M is Strontium.

<u>Explanation:</u>

Let us consider the molar mass of metal be 'x'.

The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol

It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

$\frac{15.44}{100}\times (x+16)=16g/mol\\\\(x+16)=\frac{16g/mol\times 100}{15.44}\\\\x=(103.626-16)g/mol\\\\x=87.62g/mol$

The metal atom having molar mass as 87.62/mol is Strontium.

Hence, the element represented by M is Strontium.

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3 years ago

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Define and then describe the difference between a polar covalent bond, nonpolar covalent bond and ionic bond.

Romashka [77]

Polar covalent bond- a bond where atoms are unevenly shared due to a larger difference in electronegativity of the bonded elements.

Non-polar covalent bond- These are bonds between elements with a low difference in electronegativity. Electrons are shared equally in these bonds between the elements.

Ionic bonds- have such large difference in electronegativity that they take/give electrons to the element they are bonded to. They do not share electrons at all. Bonds between a non-metal and a metal.

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4 years ago

For all of the following questions 20.00 mL of 0.200 M HBr is titrated with 0.200 M KOH.

HACTEHA [7]

Answer :

The concentration of $H^+$ before any titrant added to our starting material is 0.200 M.

The pH based on this $H^+$ ion concentration is 0.698

Explanation :

First we have to calculate the concentration of $H^+$ before any titrant is added to our starting material.

As we are given:

Concentration of HBr = 0.200 M

As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion $H^+$ and bromide ion $Br^-$ .

As, 1 M of HBr dissociates to give 1 M of $H^+$

So, 0.200 M of HBr dissociates to give 0.200 M of $H^+$

Thus, the concentration of $H^+$ before any titrant added to our starting material is 0.200 M.

Now we have to calculate the pH based on this $H^+$ ion concentration.

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (0.200)$

$pH=0.698$

Thus, the pH based on this $H^+$ ion concentration is 0.698

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3 years ago

Reaction rate is expressed in terms of changes in the concentration of reactants and products. Write a balanced equation for the

KengaRu [80]

Answer : The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

Now we have to determine the balanced equations corresponding to the following rate expressions.

$Rate=-\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{d[CO_2]}{dt}$

The balanced equations will be:

$CH_4+2O_2\rightarrow 2H_2O+CO_2$

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3 years ago