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GuDViN [60]
3 years ago
8

Calculate the speed. 1. If a car travels 800m in 30 seconds how fast is it going?

Chemistry
1 answer:
Alex787 [66]3 years ago
4 0

Answer:

s = 26.7 m/s

Explanation:

Given data:

Distance travel = 800 m

Time taken = 30 sec

Speed of car = ?

Solution:

Formula:

S = d/t

s = speed

d = distance

t = time

by putting values.

s = 800 m/ 30 s

s = 26.7 m/s

You might be interested in
How do you determine how many molecules are in a chemical equation
V125BC [204]

Answer:

Determine the mass of substance and it's molar mass. Divide the given mass by it's molar mass to get moles and multiply time

6.002*1023 molecules 1mole

5 0
2 years ago
Predict which member of each pair will be more acidic. Explain your answers. (a) methanol or tert-butyl alcohol (b) 2-chloroprop
Art [367]

Answer:

A. Methanol

B. 2-chloropropan-1-ol

C. 2,2-dichloroethanol

D. 2,2-difluoropropan-1-ol

Explanation:

Primary alcohols are stronger acids than secondary alcohols which are stronger than tertiary alcohols.

This trend is so because of the stability of the alkoxide ion formed(stabilising the base, increases the acidity). A more stabilised alkoxide ion is a weaker conjugate base (dissociation of an acid in water).

By electronic factors, When there are alkyl groups donating electrons, the density of electrons on th O- will increase a d thereby make it less stable.

By stearic factors, More alkyl group bonded to the -OH would mean the bulkier the alkoxide ion which would be harder to stabilise.

Down the group of the periodic table, basicity (metallic character) decreases as we go from F– to Cl– to Br– to I– because that negative charge is being spread out over a larger volume that is electronegativity decreases down the group.

Electronegative atoms give rise to inductive effect and a decrease in indutive effects leads to a decrease in acidity. Therefore an Increasing distance from the -OH group lsads to a decrease in acidity.

From above,

A. Methanol

B. 2-chloropropan-1-ol

C. 2,2-dichloroethanol

D. 2,2-difluoropropan-1-ol

6 0
4 years ago
The reaction P4 +502P4010 + 712 kcal is
igor_vitrenko [27]

Answer:

The reaction is exothermic (option 4)

Explanation:

P4 + 5O2 → P4O10 + 712 kcal

In chemical reactions heat can be absorbed or released:

⇒in the first case, when heat is absorbed, this is called an endothermic reaction. The products have more energy than the reactants. The reaction requires or absorbs energy from it's surroundings.  That means  that in this reaction energy , in the form of heat, will be absorbed by the reactants.

⇒ when heat is released, this is called an exothermic reaction. The reactants have more energy than the products. The reaction gives or releases energy to it's surroundings.  That means  that in this reaction energy , in the form of heat, will be released by the reactants.

in the case of P4 + 5O2 → P4O10 + 712 kcal

We notice that on the right side, which is the product side, there is a positive amount of energy. This means that the energy is released by the the reactants, in this reaction. <u>The reaction is exothermic.</u>

.

4 0
3 years ago
Read 2 more answers
Given that sodium chloride is 39.0% sodium by mass, how many grams of sodium chloride are needed to have 100.mg of Na present? G
seraphim [82]

<u>Answer:</u> The mass of sodium chloride solution present is 0.256 grams.

<u>Explanation:</u>

We are given:

39.0 % of sodium in sodium chloride solution

This means that 39.0 grams of sodium is present in 100 grams of sodium chloride solution

Mass of sodium given = 100 mg = 0.1 g     (Conversion factor:  1 g = 1000 mg)

Applying unitary method:

If 39 grams of sodium metal is present in 100 grams of sodium chloride solution

So, if 0.1 grams of sodium metal will be present in = \frac{100}{39}\times 0.1=0.256g of sodium chloride solution.

Hence, the mass of sodium chloride solution present is 0.256 grams.

8 0
3 years ago
10.0 g Cu, C Cu = 0.385 J/g°C 10.0 g Al, C Al = 0.903 J/g°C 10.0 g ethanol, Methanol = 2.42 J/g°C 10.0 g H2O, CH2O = 4.18 J/g°C
Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

Q=mc\Delte T

Q = Energy gained or lost by the substance

m = mass of the substance

c = specific heat of the substance

ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the copper = c =  0.385 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.385J/g^oC}=25.97^oC

2) 10.0 g of aluminium

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the aluminium= c =  0.903 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.903 J/g^oC}=11.07^oC

3) 10.0 g of ethanol

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the ethanol= c =  2.42 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 2.42 J/g^oC}=4.13 ^oC

4) 10.0 g of water

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c =  4.18J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c =  0.128 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

3 0
4 years ago
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