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Mashutka [201]
3 years ago
10

What is matter? 12 points..

Chemistry
1 answer:
Nadya [2.5K]3 years ago
8 0

Explanation:

matter is a substance that has mass and it takes up space.

hope this helps!!

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Calculate the volume of a liquid with a density of 5.45 g/mL and a mass of 65 g
avanturin [10]
The equation to calculate Density is Mass / Volume. You are given that the density is 5.45 and the mass is 65; 5.45 = 65 / v. So v = 65 / 5.45; v = 11.93 mL (or if you want your answer to consider significant figures, v = 12 mL).
6 0
3 years ago
During photosynthesis, the following reaction takes place: carbon dioxide + water + light energy → sugar + oxygen During cellula
Pachacha [2.7K]

Answer:

It cost Evan $17.70 to send 177 text messages. How many text messages did he send if he spent $19.10?

Explanation:

I cant do this

7 0
2 years ago
Write an equilibrium equation that shows bisulfate acting as a weak acid in water.
NemiM [27]

Answer:

Explanation:

Bisulphate ion is a weak acid as it can form hydronium ion in water .

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

The equilibrium constant of this reaction is very small , hence bisulphate ion is very weak acid.

4 0
3 years ago
The average rate of consumption of br− is 1.86×10−4 m/s over the first two minutes. what is the average rate of formation of br2
Scorpion4ik [409]
By considering the reaction equation is:
5Br(aq)+BrO3(aq)+6H(aq)= 3Br2(aq)+3H2O(l)
when the average rate of consumption of Br = 1.86x10^-4 m/s
So from the reaction equation 
5Br → 3Br2 when we measure the average rate of formation (X) during the same interval So,
∴ 1.86x10^-4/5 = X / 3
∴X = 1.1 x 10^-4 m/s
∴the average rate of formation of Br2 = 1.1x10^-4 m/s


7 0
3 years ago
In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691
andre [41]

Explanation:

Let us assume that the value of K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s

Also at 1500 K, K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s

                     K_{r} = 400.613 m^{6}/mol^{2}s

Relation between K_{p} and K_{c} is as follows.

                  K_{p} = K_{c}RT

Putting the given values into the above formula as follows.

                  K_{p} = K_{c}RT

         0.003691 = K_{c} \times 8.314 \times 1500

                K_{c} = 2.9 \times 10^{-7}

Also,     K_{c} = \frac{K_{f}}{K_{r}}

or,                K_{f} = K_{c} \times K_{r}

                               = 2.9 \times 10^{-7} \times 400.613

                               = 1.1617 \times 10^{-4} m^{6}/mol^{2}s

Thus, we can conclude that the value of K_{f} is 1.1617 \times 10^{-4} m^{6}/mol^{2}s.

6 0
3 years ago
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